Archived questions and answers for 2020, Sp 00

Extra Help opportunities in 2020?

There will be some evening help sessions, but I don't have any specific
times or places yet. All this information, including all the regular TA
office hours, is available on my  Office
hours/help  link (available on the main page, under "Quick Links")
The physics office keeps a list of  private tutors  (who
charge money), but I don't know anything about those individuals, so
can't make any recommendations.  

Capa question from someone who hasn't used it before:

> I was wondering  could I look at the CAPA on the internet and not plug
>in any answers and still have the same amount of tries? I am new to
>this.

You bet. You can log in to CAPA and look at it, and in fact you can
fill in SOME boxes and not others if you want. When you hit any "submit
all answers" box, it ONLY looks at the boxes you filled in. So it won't
take away any "tries" from the ones you didn't explicitly try. (Just be
careful to avoid hitting the "refresh" button on your browser, because
if there are any boxes with old tries in them, that'll eat up a try on
those problems)  By the way, if you get the answer wrong because of
"units", that does not count as a try.

So, you can look at it, log out and in as many times as you want, and
nothing counts as a try except explicitly filling in something for that
problem and hitting a submit button.

By the way, I usually give about a dozen tries on any problem, so you
really shouldn't run out of tries.  (You won't get CAPA's by guessing,
usually!) If you ever get close (down to 1 or 2 left) I strongly
suggest you come see me, and figure out what you're doing wrong, before
using up the last ones.

Hope this helps, thanks for writing. See you soon -

Question on CAPA #1, Problem 6

 Someone wrote with all their calculations: they properly found the
distance "r" to each charge by dividing the distance by two, (since
you're in the middle) They had the correct formula for "E" = kQ/r^2.
They had computed "E" from the left hand charge (which came out
negative, because their lefthand Q was negative) and "E" from the right
hand charge (which came out positive, since their righthand Q was
positive) And finally they added these two numbers... and concluded:
>CAPA tells me this is wrong, but I don't see where I made my mistake...  


Almost everything you did is correct, but there's one little problem
that gave you the wrong answer. It has to do with *signs*.  Remember in
class I suggested you use Coulombs law to figure out the MAGNITUDE of
forces, but then look at your picture and put all the minus signs in by
hand, to make sure you get them right.  (That's just as true for "E" as
for "Force", because "E" is also a vector quantity)  Just plugging in
numbers and charges into Coulomb's law can get you into some trouble
with signs (that's in fact the  main point of this problem!)

You had two quantities to add, the first one came out negative and the
second came out positive. But draw a little picture for yourself. I
think you should be quick to convince yourself that BOTH E fields point
to the left, i.e. both terms should be negative! (The minus charge
attracts you to the left, the positive charge pushes you to the left,
BOTH point left!) If this doesn't make sense, stop by my office, or
after class, a quick picture on the board should be convincing.

Question on Set #1, #4

>       I can't seem to do numbet 4 on the CAPA.  I've worked on it for a
>couple of hours now and am really frustrated.  I think that I am using the
>equation E=F/q right?  Yet, when I plug the numbers in from my calculator
>to the CAPA, they are wrong. I don't understand what I am doing wrong and
>I think that I"ve tried everything that I could think of.  Please help.


In general, it would help if you told me a little more. You gave me the
formula (good!) but ALSO telling me your numbers, and answer, might
help.  In this case, I looked at what you'd been submitting to CAPA,
and it seems that you just rounded a little too much! You need to get
the answer right to within 2% for CAPA to know you got it. So if, say,
the true answer was 1.04, and you entered 1.0, you'd be off by 4%, CAPA
would simply say "no"!  So go back and use one more digit.

CAPA gave 3 sig figs in all the numbers for this problem, and in
general (even if it DOESN'T) it's probably wisest for YOU to keep (at
least) 3 sig figs thoughout, because rounding to 2 digits can easily be
more than a 2% roundoff error...  Make sense?

Good luck!

Question on Set 1, #8

>I am a little confused by the hint given after #8.  How are the
angles going to help?

You have to pick a charge and find the TOTAL force on it (that's the
problem.) Suppose you pick the charge at the top of the triangle. There
are two other charges, you might call them "1" and "2".  So you need to
find F1, and F2, the force arising from charge 1, and the force arising
from charge 2, and add them.

Seems easy, because the force (F1, or F2)  is obtained by Coulomb's law
(you know the charges, and the distance between them is given!). But
there's a small problem - yes, you do know the magnitude of the two
forces by Coulomb's law, but alas you can't simply add those numbers.
You can't because F1 and F2 are *vectors*, they're tipped! So you need
to draw a little force diagram, with F1 and F2. You get their LENGTHS
by Coulomb, and you get their DIRECTION by just looking at the picture,
and drawing carefully. That's why my hint mentioned the angles: if you
know the angles in the equilateral triangle, you can immediately write
down the angle those vectors make with the horizontal. 

From there, it's just a vector addition problem. (So, you find the x
and y components, add them up,.... the usual old story. Here, there is some
"symmetry" that makes the calculations pretty quick..) 
Does that help you get going?

A question about the material

>In yesterday's notes in the beginning of the field line discussion you
>said a +Q has field lines radiating and pointing outwards from +Q.  On a
>single charge like +Q do the lines move out to infinity?  I have it in my
>notes, but you don't in yours.

If I understand the question, yes, field lines run all the way off to
infinity. (Either that, or they run into another charge and end there.)
For a point charge +Q, the lines just run off to infinity. If there was
a point charge +Q off in another galaxy, we would feel its attraction
(if we use a positive test charge, anyway - or repulsion, if we use a
negative test charge). Of course, the *strength* of attraction, by
Coulomb's law, goes like 1/r^2, so it would be quite tiny if r is big.

Now you're leading me off on a tangent:  Gravity is the same way.
Gravitational attraction runs all the way off to infinity, getting
weaker and weaker like the square of distance, but it never "cuts off".
We feel the pull of very distant galaxies. In that case, gravity wins
because distant  galaxies have lots of mass, but little or zero net
charge.(most matter is electrically neutral.  In fact, it's a
plausible hypothesis that the universe as a whole is overall
electrically neutral.)

I didn't talk in lecture about "gravitational fields", but they work
the same was as E fields. E = F/q, it's the "unit electrical force per
charge". Similarly, you can define a gravitational field = F/m, the
"unit gravitational force per mass". Near the earth, the field is F/m =
mg/m = g, it's a UNIFORM (constant) gravitational field which points
down, everywhere!  It makes a nice analogy to think about. There is a
gravitational field everywhere in this room. It points down, and has a
constant value of 9.8 m/s^2. (The units of gravitational field are
different from electric field, because we divide by something different
in the definition)   It is present EVEN if there is no object there.
It is present right in the middle of other objects too!

E fields are like that too, but of course they aren't in general
uniform. E.g, E in between two charged plates (inside a capacitor), as
we saw in the demo last lecture, IS uniform, so it's a lot like a
gravitational field.  But instead of pointing towards the earth, it
points towards the negative plate...

Sorry about the digression, hope that answered your question - 

Demo question

>Just a quick question.  I noticed a smell in the air after the 
demonstration with the Tesla Coil.  What was the smell due to?

I believe the smell was ozone. High voltage breaks down the oxygen
molecules (O2) in the air, leaving charged "free radicals" of Oxygen
which can attack other O2 molecules, giving you O3 (ozone). It is
pretty short lived (an unstable molecule) and has a characteristic
smell. (Can you think of times/places where you've smelled that before?
Was it near a high voltage source?)

Of course, it also might have been burnt professor. That Tesla coil
smarts a little when you get close to it!

A Giancoli end-of-chapter problem

 >Prob # 3 (Chapt. 16) - I understand that Fe goes like 1/r^2. So if you
>triple the F then you (1/3)^2 the distance, right? 

No, if you triple the F, then you (1/3)^0.5 (square root of 1/3) the
distance!  (I.e, multiply 20 cm by Sqrt[1/3] to get the new distance)

The reason is that after you SQUARE that new distance, it should to
make F go up by 3.

Here's a mathematical way to see it:

Remember the original F = kQQ/r^2. 

Now we change r to r', and F to F'=3F (we triple the F, and we're after
the new distance, r')

So F  = kQQ/r^2 and 
   F' = kQQ/r'^2

Let's divide those equations: (The k's and Q's all cancel out)

F/F' = r'^2/r^2  (Do you see why the prime on "r" is upstairs?)

1/3 = r'^2/r^2, now take the square root of both sides

Sqrt[1/3] = r'/r.  The new radius is Sqrt[1/3]*r.  
They gave r, so just multiply by Sqrt[1/3] to get the new radius. 

-----------

Maybe it would've been easier if they'd said the force was QUADRUPLED.
Then, can you see that the radius must be *half* what it was? (Because if
radius is half, then radius^2 is half^2 = 1/4, and that's downstairs,
so the force is 4 times bigger.

A question about Giancoli, P. 506 - and a long winded answer!

>2. Pge 506 bottom paragraph - When E is not uniform.  I don't understand why
>delta V needs to be negative.  I think I am still confused about the
>direction of V - I understand that V = PE/q so that V is independent of the
>test charge - should I use this to determine the sign of V?  Or should I
>think about the direction that the charge moves - that + to 1 would be +V
>and - to + would be -V? I don't know why I am so confused about this....  

I don't quite understand the wording of your question. Let me try to
answer what I *think* you might be asking :-) If I missed the point,
try again...

First, potential (V) is always HIGHER near + things and LOWER near -
things. 

To understand that, imagine putting a + test charge down near the
things.  Remember, PE = qV, that's the fundamental definition. So if q
(my test charge) is +, then PE and V must have the *same sign*.
(Multiplying V by a positive q doesn't change the sign) If you are +
and you're near a + thing, you have lots of (positive) potential
energy. (You WANT to run away to infinity, to zero V) That's why V is +
near + things.  But if you are a + test charge and you're near a -
thing, you have NEGATIVE PE (someone would have to do work on you, ADD
energy, to bring you to zero energy at infinity) so that's why V is -
near - things.

Actually, I should be careful: V is simply LOWER near - things and
HIGHER near + things. You can DEFINE V to be zero wherever you want (If
I define it zero at infinity, all the things I said above are fine. But
I sometimes define it to be zero near the negative plate of a
capacitor.  So V is 0 near a negative thing! Is that contradictory with
the above? No - just so long as you see that the + plate of the
capacitor must then be at a HIGHER voltage)

[Digression: calling V=0 at the negative plate can seem strange. That
means PE=0 at that plate, EVEN for a negative test charge. But that
negative test charge WANTS to fly away, towards the positive plate, and
do work. How can that be, if it has PE=0 to start with? The answer is
that when it gets to the + plate, it has PE = qV =
(negative)*(positive) = negative, it has LESS PE than it started with!
All that matters for objects to spontaneously go flying away is that
they lower their PE, it doesn't matter if you call the starting point 0
or not)]

Anyway think about what we've concluded: V is always HIGHER near +'s,
and LOWER near -'s. But electric fields ALWAYS point away from +'s, and
towards -'s. So Electric fields ALWAYS point away from higher V towards
lower V!!

Now we get to Giancoli's story on P. 506:  Suppose you have two points
a and b, with a on the left and b on the right, with voltages Va and
Vb. Let's call Delta V = Vb-Va.  Imagine first that delta V>0, i.e.  Vb
is HIGHER.  Remember, that means the electric field points AWAY from b
towards a, i.e. it points from High V to low V, or in this case RIGHT
to LEFT, i.e E is negative.  Summarizing: when Delta V is >0, E<0.
(That's why Giancoli says E = -Delta V / Delta x, that's the minus sign
I'm worrying about) 

Imagine instead that delta V<0.  That means Vb-Va<0, or Vb is at LOWER
potential, Va is higher now.  That means the electric field points from
a towards b (E field always points from high potential to low, that's
the way + charges WANT to go!)  So in this case, E points from a to b,
i.e. to the right, i.e.  E is positive. So Delta V<0 means E>0.  Again,
E has the *opposite* sign of delta V. I think that's what Giancoli was
trying to explain on P 506.

(The fact that its E = - Delta V/Delta x he hasn't really tried to
exlain, but it's just from the definiton of work.  *Very* sketchy:
W=F*delta x. Now replace W with Delta PE (work = change in energy) and
then replace Delta Pe = q Delta V (our basic definition of V) on the
left. Replace F with q E on the right, and you get Delta V = E Delta x.
The minus sign is from the arguments we just made!)

I have a rule of thumb about signs: 

For Force and E problems, use the formulas to give you *magnitudes*
(take the absolute value of ALL charges, at all times!) Then look at
the picture, and put in all minus signs by *hand*. The minus signs in Force
and E problems have to do with DIRECTIONS, so do NOT let the formulas
trick you into putting in signs based on the signs of the "q"s
involved... 

ON THE OTHER HAND, the signs in "V" and "PE" problems have nothing to
do with direction, so you must let the formulas put in the signs!  Just
use the formula! If PE = qV, and q is negative and V is positive, then
the formula says PE is negative, and that's right. Don't try to second
guess the signs in the energy problems (or voltages), let the equations
guide you.  Afterwards, you might be able to think about the sign and
convince yourself it's reasonable (like, think about whether the system
wants to DO work to get to zero PE, or whether you'd have to do work ON
it), but that's just a CHECK, you shouldn't stick any minus signs in by
hand in PE or V problems...

What are eV's?

>3. Page 508 - The last paragraph talks about electrons moving through
>different potential differences - I don't understand how you calculate the
>eV.  This is probably really obvious and I am just not seeing it. Sorry.

We'll do this in class on Wed. There are two ways to deal with eV's.
The "brute force" way is just to realize that eV is a unit, the
conversion is 1 eV = 1.6E-19 J.  (eV is a unit of ENERGY. It's just the
NAME of the unit.  You have "Joules" or "eV"'s. ) That's it. Using eV's
is no harder than replacing cm with inches, using 1 in = 2.54 cm.

The nice thing about eV's is to remember PE= qV.  So if "q" (the
charge) happens to EQUAL e (i.e. if you have a proton), and if V is in
voltes, then PE = e*V comes out in eV automatically. 

Example:  A proton, starting at rest, moves through 10 V of potential
drop. What is the final kinetic energy? 
(The formula is just q Delta V.  Loss in PE in this case is increase in KE)

So the answer is KEf = q*V = (e)*(10 V).  
You can plug in e=1.6E-19 C if you want, giving KEf = 1.6E-18 J.  
THEN you could convert to eV, 1.6E-18J * (1eV/1.6E-19J) = 10 eV.  
Lots of calculator work! 

Look at that final result symbolically:  
KEf = (e)*(10V) = (1.6E-19 C)*(10V) * (1 eV/1.6E-19 J).  
See how the number "1.6E-19" which is the CHARGE of e
cancels nicely with the "1.6E-19" which is the eV to J conversion.
(That's why we PICKED that particular conversion: so those numbers would
cancel) leaving behind just the NUMBER of volts.  (KEf = 10 eV)

So look: (e)*(10V) = 10 eV!  Simple, no? 

A proton dropping 20 V gains  (e)*(20V) = 20 eV of energy.

An electron going through 20 V of potential change gains 20 eV of energy too
(it goes the other way, it goes UP in potential, but the charge is the same
magnitude as the proton, so the energy comes out the same)

An alpha particle has charge 2e. If it moves through 10 V, it gains
(2e)*(10V) = 20 eV. 

That's why eV's are nice, if the charge is e (or a simple multiple of
e), the *arithmetic* is easier using "eV"'s as your energy unit, that's
all.

Old exams to study

>I was just wondering if there is an exam on the web that I may be able to
>take a look at in order to help me study.  I looked on the homepage in the
>places that I thought I might find one, but to no avail.  Can you help
>lead me in the right direction?

Since this is the first time I've taught 2020, I don't have any old
exams of my own for you to use to study. However, you might look at my
 "Other links"  in the bottom right
corner of the web site.

(About halfway down *that* page are links to "Old course pages":  there
are exams on all those sites to study with.)

CAPA #2 , problem 4

>So, on number 4 we have an electron starting from rest a certain =
>distance away from a fixed point charge (with some given charge) and we =
>are asked how fast the electron will be moving when it is very far away. =
> It's that 'very far away' part that has me confused.  Is the distance =
>irrelevant?

The point is that "very far away" means "the potential energy is
basically zero".  Initially, potential energy is high, Force is high,
and acceleration is large. The farther it gets, the smaller the
acceleration (but still positive, so it IS still speeding up) Far away
means that the potential energy has become negligible, so all its
initial energy (which was pure PE) has converted into kinetic.  (In the
formula, PEf = k Q/r(final), so if r(final) is HUGE, PEf is basically
zero...)

CAPA #2, Prob 5

> On number five, I believe it's quite similar to 4 but I'm still =
>running into some problems.  I used that same method of setting KE=3D =
>-PE but I don't think it's right because there's a given distance =
>between the two plates and I don't know where it fits in!?

The rule of thumb here should be "conservation of energy". So the
formula you want is KEi + PEi = KEf + PEf... 
(Initial energy = final energy)

In Problem 5, KEi = 0, KEf is what you're after. Solving for it gives
KEf = PEi - PEf.  What is PEi - PEf here? The plates have a given DIFFERENCE
say 180V. They didn't say which side is higher, but you know that - the
positive plate is always higher. Since an alpha particle runs from
positive to negative, (i.e. from high voltage to low) PEi = +q*Vi =
(2e)*180V.  (With my usual conventions, PEf = 0)  Does this help?

Set 2, #10

>And the last question is about #10...if the E field is constant and 
>equals 77V/m (in my question) and I'm given that the potential at point 
>1 is 1000V, then by definition shouldn't the distance be 12.98m? 
>( E=V/d )

So it seems you're taking E=V/d and turning it around, to get d = V/E =
12.98 I guess?  No, that's not right... You have to think more
carefully about the formula E=V/d, (or, if you prefer, V=Ed). That's
not a *fundamental* formula, we derived it in the context of a very
specific problem: What is the voltage across a capacitor of width d?
Really, what I had on the board was V = Ey,  and what that MEANT was
that as you go from y=0 (the negative plate) towards y=d (the positive
plate), the voltage is steadily (linearly) increasing. Look at Fig 17-5
which shows some "green" equipotentials, THAT is really what the
formula is trying to tell you about. As you move from one plate to the
other, the value of the potential is steadily changing, with 
Delta V = E*delta y.  
That's a slightly more general (and fundamental) formula!  (Though, it only
works perfectly if you're in a capacitor, or otherwise if delta y is small)

Maybe if you use IT, you can see better how to work this problem?   The
answer is, in the end, fairly straightforward. This one is easier to
explain at a blackboard - can you stop by after class on Friday?

Good luck -

Review session?

(I was asked about my review session - I hold an extra office hour Monday
before exams, 6-7 PM in G1B30.  A grad student will be there for another
hour to continue with questions) 

assorted questions from someone

>Regarding CAPA #1, Problem #6,
>I am unclear why both E fields are negative since only one charge was
>negative. 

It's because the DIRECTION of E depends on *two* things: the sign of
the charge, AND where you are in relation to it.  The sign of E tells
you which way the E vector points. (My convention is usually "left" is
negative and "right" is positive on the x-axis)

If you are a + test charge to the RIGHT of a negative charge (which you
are in this problem), it attracts. In other words, the E field due to
that negative charge on your left is towards the left, or NEGATIVE.

On the other hand, if you are a + test charge to the LEFT of a positive
charge (which you again are, in this problem), it repels. In other
words, the E field due to that positive charge on your right is again
to the left, i.e. again NEGATIVE

So both E-fields point to the left, and THAT is where the minus signs
come from. 

>On Capa Set 1, problem 11. The solutions state that "Where the E field
>lines are denser the E field must be weaker." is a true statement. I got 
>this right by choosing false which is what I thought was correct.Did I 
>miss something?

Yikes, no, my mistake. Just a typo, I just fixed it. Thanks!

>I am unclear about the relation of the E field to potential difference. 
>Can you help? 

It's not something I expect people to know quantitatively, but
qualitatively, it's kind of nice to have a feel for it...  Remember, in
a capacitor, E*d = V. (This formula is not true anywhere else besides
capacitors..)  However, if you write that formula a little more
carefully, like this:

Delta V = E * delta x,

it turns out that IS always true, as long as delta x is small.  (In a
capacitor, delta x doesn't HAVE to be small for the formula to still
work). So that's the relation of E field to potential difference.
E = Delta V / Delta x.

Think of contour maps. If you have lots of contour lines very closely
spaced, that usually means: cliff!! (You get a LOT of elevation CHANGE
over a very small horizontal distance) And cliff means you feel the
force of gravity maximally! So where the V contour lines are closely
spaced, the force (E) is big. When contours lines are far apart (not
much change in voltage going on) there's not so much of an E field.

I think that's really all you need to have a feel for.  Also the sign
of E is worth understanding (E always points from HIGH V towards  LOW
V. Always. It's like gravity always points from high height to lower
heights)

Example: When I approach the Van de Graff, I don't generally get
sparked until I'm CLOSE: the Van de Graff is at some voltage, I'm at
zero, and when Delta V/Delta x gets big (i.e. when I'm close, since
Delta V is some fixed number) the E field gets large enough that F=qE
can start to rip apart air molecules, there's a spark, and I get
zapped.

A random physics question:

> I am in your MWF 11:00 class.  I was =
>wondering if you could explain how a black light works.  Like why it =
>makes white objects bright, especially bleached clothes.  I thought it =
>had something to do with the fact that white reflects every color, so in =
>a dark room with a black light, the white objects are illuminated.  =
>Also, do you know what a black light is made of?  Last, are black lights =
>harmful to you?

I have some ideas about black lights, but I went to a wonderful web
site (which I even have linked from our web page) called "How things
work", just to check.  The site is 
http://rabi.phys.virginia.edu/HTW/ , and you can search on it for
LOTS of cool questions about everyday physics phenomena.

We're going to talk about light in about a month in 2020, but the quick
answer is that "black lights" are bulbs emitting light that's a little
higher energy (shorter wavelength) than the normal visible light from
ordinary bulbs. (It's called "ultraviolet", or UV light - the same
radiation that causes you to tan, or burn, in the sun )  Your eyes
can't directly detect it (that's why it's "black"), but there's still
plenty of electromagnetic radiation leaving the bulb. Certain materials
(I believe it may be the chemicals in the bleach, rather than the white
cotton itself) absorb this UV light, which excites the atoms, and then
when they "de-excite" they emit a cascade of lower energy photons
(light) that you can see, hence the glow. Some materials do this,
others don't, that's why some objects appear to glow under the lamp and
others don't. I don't *think* it has so much to do with whether the
object is originally white or not. But I'm not sure. You could do an
experiment (wash an old dark colored shirt with bleach, and see if you
can pick up any glow under the black light?) As for harm, I'm not sure.
I guess if it's low powered, it's probably fine, but I wouldn't just
leave it on all the time. (But then, I use factor 30 sunscreen when I'm
outside in the sun!) 

CAPA 3, #2

> ... I don't know what gauge is 

Neither do I, but that's o.k., because they TELL you the diameter (and that
is what gauge is some kind of measure of)

Computer room, password

 Does our Physics 2020 class have a code to use the computer room
that is next to the physics library?

You bet. The username is superatom, password is kx8qzt

CAPA #3, Problem 3

>    #3--  The picture shows the schematic of a battery and two
>resistors.  I'm curious as to why part C ("The voltage drop across wire
>B is larger than that of wire A," where wire B has greater resistance
>value) is false.  It seems that the "voltage drop" across the wire with
>the largest resistance will be greatest, no? Perhaps I am unclear on
>what "voltage drop across a wire" really means.

Look at it this way. Suppose I call the voltage on the ideal wire at
the bottom of the picture "0 V".(It's the same everywhere along an
ideal wire, right?)  Then look on the left: there's a battery with SOME
voltage V, which means the ideal wire at the TOP of the picture is
everywhere "V", all along it!  So what's the "voltage drop" across
either resistor?  It's the voltage at the TOP of the resistor minus the
voltage at the bottom, i.e. V-0 = V, the SAME for both of them!

You said the voltage drop across the wire with the largest R will be
greatest, but that would ONLY be true if the CURRENT was the same
through them! But there's no reason why it should be (on the contrary,
since V=IR, and I've just tried to convince you that V is the SAME
across either resistor, this tells you current will be LESS through the
larger resistor!) Make sense?

>Since the wire of A is twice the diameter of B, The area of wire 
>A is twice the area of B, correct? Therefore, the resistance of B is 
>twice the resistance of A, correct? 

No, area goes like diameter squared! (Area = Pi R^2 = Pi D^2/4)

> I am unclear about what "Power dissipated" means. 
"Power dissipated" is in Giancoli 18-6. There are three formulas, ALL 
EQUIVALENT, for Power dissipated in a resistor:
P=IV
P=V^2/R
P=R*I^2

(They are all equivalent because V=IR)  

V here means "the voltage across the resistor", i.e. V(one side) -
V(other side).  Based on the discussion above, you know that it's just
"V", the *same* for both resistors. (That won't ALWAYS be true, of
course, it depends on the particular setup!) "I" means the current
going through THAT particular resistor.

You have to think about which formula to use. Since we decided (above)
that the current is NOT the same through the two resistors, I suggest
you AVOID a formula using "I" in it...

Problem #7

>    #7-- I know I'm not the only one baffled by this one.  It seems that
>if you plug in the numbers for the initial voltage and wattage, you'll
>find the current.  Plug in this current value to P=IV with a new voltage
>and find the new wattage, right?  Well, I'm stumped.  I may be confusing
>definitions here as well.  Is the "wattage rating," different than
>electrical power watts?

(Wattage rating is indeed just Power) We really didn't have a chance to
get to this in class today, my apologies!  Yes, plugging in initial V
and Power gives you current, but that's not what you really want. After
all, that's the current through the bulb in Europe. But, there's no
reason to expect the CURRENT will be the same in the US! Think about it
- when you plug something in here, the voltage is different, so the
current will be too! What's the same is the piece of metal in that
bulb, the RESISTANCE R is what stays the same no matter what country
you're in!  Does that halp you get going?

CAPA 3, #4

>I have a Question about CAPA 3,  #4.  Should I be thinking, What
>fraction of  the total length is the short wire segment?  

Yes, I think this is an ok way to be thinking about the problem.  (I
think more about how much bigger the long segment is than the short
wire segment, like Long = 10*short, or whatever that number works out 
to be...)

>I have been
>following the conceptual example 18-5 on Pg. 535 in Giancoli and so far
>that proportion hasn't led me to the right answer.  Should I not be
>thinking of this this problem in terms of V=AL???


No, that conceptual question is quite different!! There, someone took
some wire and stretched it, making it longer but skinnier (keeping
volume fixed). That is NOT what happens ordinarily, certainly not if
you lay down wires underground - you never *stretch* the wire (that
would surely break all the time!) It's always the SAME exact kind of
wire everywhere (with the SAME cross sectional area everywhere), it's
just that you have different lengths.  For identical wire, R is simply
proportional to LENGTH, and you know all the relevent lengths in this
problem, I think it's pretty straightforward! (?)

CAPA 4, #7

 
> I was wondering if you could explain why you use the Vzero measurement in
>problem 7,  and not the Vrms.  Do voltmeters measure the Vzero?  Or is it
>arbitrary-- can they measure either?  I initially converted the Vzero to
>Vrms for this problem, but didn't get the right answer.

If I understand correctly, you're assuming V0 is AC? But it's not,  in
problem 7 (as in most problems on this set), the voltage source is just
a DC battery. (The symbol for AC voltage is a little wiggle in a
circle, the symbol for a DC battery is a long and short line) There's
no such thing as "V_rms" for a battery. (Or if you prefer, there's no
DIFFERENCE between V_rms and V0 for a battery - if V0 is not changing
with time, the average and instantaneous values are always the same)

(By the way - a real voltmeter CAN be set to read either AC or DC, and
when you set it to read AC, it does read V_rms.)

CAPA 4, #6

>... After connecting a and b to a battery, the voltage across R1 always 
>equals the voltage across R2....  Your hint told me that ..

Hmm, perhaps my hint may be a little misleading for this part. (It was
meant to clarify the "less than OR EQUAL TO" in some of the *other*
parts: I didn't want people getting picky on me and arguing that it has
to be STRICLTY less than (or greater, whatever) rather than less than
OR equal to.... ) So for the question you're describing, you should NOT
assume either resistance might be zero. It really can't be - if you try
to put *zero* resistance across a battery, you get INFINITE current,
which means sparks and light and dead batteries and... well, just don't
short out batteries, please! The point here is that when resistors are
in parallel, the voltage drop MUST be the same across each of them.
(Look at the picture! If you have voltage "Va" on the  left, and
voltage "Vb" on the right, *either* resistor has the *same* delta V =
Va - Vb across it!!) Series resistors have the same CURRENT flowing through
them, parallel resistors have the same VOLTAGE across them. Look at the
pictures and convince yourself of these facts!

A common problem with sig figs - please read!

 

The details are different each time, but here's a very common problem
people are having: you're sure you've done the work correctly, but CAPA
still says NO. The reason is often that many of you are rounding your
answers off too much!  E.g, suppose you calculate an answer of, say,
.00124, but you input .001,  or even 0.0012.  The first one (.001) is
19% off  (!!!) , the second one (.0012) is still 3.2% off! CAPA
(usually) does NOT care about sig figs any more, but what it DOES care
about is that you get the answer right! And "right" generally means
correct to within 1 or 2%. There's no NEED to round to two sig figs,
and in most problems its WRONG to do so.  I've been trying to be very
careful about this - I almost always give all numbers to 3 sig figs in
the problem, so your answer should too. Even if I slip up and only give
you 1 or 2 sig figs, I strongly suggest you ASSUME I meant for there to
be 3. If you round to 3 sig figs, you should be correct to better than
1%, but rounding to two digits can change your answer by TOO much!

Bottom line - keep the 3rd sig fig, and you should be fine.

CAPA 4, #5

>In this statement: The current through the resistor in circuit A is
>the same as any single resistor in circuit C, do I 
>"A)" Add the resistors in parallel to form a single resistor and
>then proceed to calculate the current flowing through it, or 
>"B)" Look at the current flowing before the resistors are grouped 
> together by using the current divider method.

If you did "A", adding the resistors in parallel to form a single
resistor and then calculating the current flowing through it, you would
be figuring out the TOTAL CURRENT coming out of the battery, right? But
that's not what the question is asking, it wants to know about the
current through "any single resistor in circuit C". (The current
through a SINGLE resistor in C is clearly smaller than the total
current, in fact, since both R's are identical here, it's exactly 1/2. Can
you see why?)

So your procedure B looks better to me! Does it make sense why I'm
saying this? It depends on the question I asked! Here, I'm asking about
the current through the *individual* resistors.

A random question

>hi dr. pollock, i just had a random question that i just know you'll be
=
>able to answer.  i was wondering how those little testers on batteries =
>work, you know, like on an energizer battery there's a place to put your =
>fingers and if it's charged up a little "good" sign appears in a little =
>window...what's the story behind those

I don't know how those things work. We just had a nice lunctime "flame
session" about it here at the nuclear physics lab, and came up with a
plausible *idea*, but I have no idea if its right.

I know there are chemicals that change color at different
temperatures.  (Dunno how THAT works, that's another question!) So
suppose you had a variable thickness in that strip, then the resistance
is different as you move along it. Hook the battery to the strip:
there's a V across it, and thus an I flowing through it. (Make the
total resistance big, so you don't have MUCH current) "I" is the same
at all points, but then I^2 R will be different as you move along the
strip, (since I'm postulating that R depends on position, e.g. by
having it taper in thickness, or width) so the local heating will be
different at different points, and so the temperature sensitive
chemicals will be different at different points. As the battery gets
older, the "V" it provides will be less, so the temperature
distribution will look different.  

I haven't ever played with one of those things, it might be fun to
bring one in and we could mess with it after class. (We can see if the
thickness indeed varies, if it is sensitive to direct heating, what its
resistance is, etc!)

NOTE: After I wrote this, someone emailed me the following. Looks like
I guessed the idea about right, but with a few details wrong!

>Thank you for visiting Energizer's Web site and also for your inquiry.
>
>The Energizer on-battery tester is a device which uses heat
>sensitive (thermochromic) material to indicate remaining power
>in the battery.
>
>Thermochromic materials change color when exposed to heat.
>In the Energizer on-battery tester, the thermochromic material
>changes from black to clear.
>
>To generate heat, the Energizer on-battery tester uses a small
>amount of electrical current from the battery.  The heat that is
>generated flows outward through the label.  This heat causes
>the thermochromic ink to change color, revealing the graphic
>layer below. 

(So, it's SIMPLER than I thought. The resistive part is probably just
localized, so all the heating happens in roughly one spot, and then the
heat flows outwards from there.  SJP)


 >Because the tester operates using heat, it is best to use it
>at room temperature.  That way, the tester is not too hot or
>too cold.
>
>To activate the tester, the consumer simply presses on the
>two green dots.  Pressing the dots closes two switches and
>connects the positive and negative battery terminals through
>the Heater Circuit.  Using the battery tester does not deplete
>battery energy.

(Hmmm.. any current drain *must* deplete battery energy. However, if
"R" is very large, "I" might be quite negligibly small, so we'll give
them this :-) SJP


>Hopefully, this information will help you to better understand
>our on-battery tester.  Once again, thank you for visiting our
>Web site.
>
>Heather
>Energizer Consumer Relations 

Capa 5, #5

>I understand Kirchoff's rules for voltage and current, but I can't seem to 
>be doing the right thing.  
>For number five, I can find the current through each of the
>resistors, but I can't get the correct current through the battery.  For the
>current at the intersection directly above E1, I should have an I coming in
>from the left, a current going out to the right, and a current going down to
>the battery.  Therefore, the current through the battery should be the
>current on the left minus the current on the right, according to Kirchoff's
>first rule.  But that doesn't give me the correct answer. 

It sounds like you do have the right idea, but I wonder if any "minus"
signs are giving you troubles. E.g, if the current "going out to the
right" turned out to be negative, you'd need to know that! (Because
then, when you subtract the current on the right, you'd be subtracting
a negative number!)  Without seeing more details of your calculation,
I'm not sure how to help any more - so my first reaction is, check all
the signs of all the currents...

CAPA 5, #9

> One question was on number nine...
> It asks how long it will take for the current in the circuit to =
>drop to .37  So I figured I would use the equation I= I (initial) =
>times e^(-t/RC). I plugged in .37 as I and I initial as the number I =
>found in #8 which was 9.257E-6. And I used R=3.5E6 (3.5Mega ohms) and =
>C=4.2E-6. Then when I solved it for t I got -155.76 sec. Did  I do =
>something wrong?

I think the problem is that I is not 0.37 Amps, it is 0.37 TIMES
I_initial.  (That is, you're being asked for the time when the current
has dropped, not to 0.37 Amps, but to 0.37* where you started, 37% of
where you started) That should help! (Negative time doesn't really make
much sense for the answer, I guess)

I have many questions...

>1. Is Pmax (max power) = Vmax * I max and Avg Power = 1/2(Vmax * I
max).>It's the Pmax that I have in my notes, but not in yours. 

Yes, P_max = V_max * I_max, and yes, Ave Power = (1/2) P_max.

 >2. Why is the resistance of a higher wattage bulb smaller than a lower
>wattage bulb? I understand to how to look at it mathematically, but 
> I still don't understand why...

"Wattage" means how much power is consumed IF you hook it up to 120 V.
Since V=IR, if you have bulbs with different R's, and hook them all up
to a 120V source, they will draw different currents. (The BIGGER R is,
the LESS current it will draw. Make sense? There is more resistance, so
you get less current) But if you have only one bulb, Power = I*V, and
V=120V is fixed, so the less current you draw, the less power you
consume... 

So High watt bulbs have small resistors, so they draw LOTS of current,
so they have a high power... Make sense? (The math says P = V^2/R, and
since V is by definition 120V for all bulbs, the smaller R, the more
power...)


>3. Example 19.7 part c. I don't get why we would want to look at the voltage
>drop accross the 5 ohm resistor.

I often find it helpful to label "points" in my picture. To find the
current through the 8.7 Ohm resistor you really just need to know the
voltage across it (and then Ohm's law will tell you the current). So if
you label the heavy dot on the left of the parallel circuit in Fig
19-10c "A", and the heavy dot on the right of that same parallel
circuit "B", what you're really after is V_A - V_B.  That's the point -
voltage V_B is NOT zero, it's "behind" that 5 Ohm resistor, so you need
to find the voltage drop across that 5 Ohm resistor to FIGURE out V_B.
Similarly, V_A is NOT 9V, because it's "behind" that 0.5 Ohm resisitor,
so you needed to find the voltage drop across the 0.5 Ohm resistor
too. (This would be easier to answer with a blackboard...)

>4. Conceptual example 19-5 part B. I don't get it.  Actually a lot of my
>questions revolve around this whole issue of who is brighter when.....

We should go over this, then. Questions like this will almost certainly
appear on the next exam! A circuit a lot like this one was in the lab,
(except A and B weren't identical, and only one was actually a light
bulb). Did that make sense to you? I think "real" office hours would be
much better for this kind of discussion...

MISPRINT ON CAPA 6, Problem 5: Please read!

A common problem with sig figs - please read!

 

The details are different each time, but here's a very common problem
people are having: you're sure you've done the work correctly, but CAPA
still says NO. The reason is often that many of you are rounding your
answers off too much!  E.g, suppose you calculate an answer of, say,
.00124, but you input .001,  or even 0.0012.  The first one (.001) is
19% off  (!!!) , the second one (.0012) is still 3.2% off! CAPA
(usually) does NOT care about sig figs any more, but what it DOES care
about is that you get the answer right! And "right" generally means
correct to within 1 or 2%. There's no NEED to round to two sig figs,
and in most problems its WRONG to do so.  I've been trying to be very
careful about this - I almost always give all numbers to 3 sig figs in
the problem, so your answer should too. Even if I slip up and only give
you 1 or 2 sig figs, I strongly suggest you ASSUME I meant for there to
be 3. If you round to 3 sig figs, you should be correct to better than
1%, but rounding to two digits can change your answer by TOO much!

Bottom line - keep the 3rd sig fig, and you should be fine.

Anonymous comments to my web site

CAPA #6, Problem 10

>Density is mass/volume, right?
>My density is 8.9E03 kg/m^3
>To find mass I need the volume of the wire, right?
>Volume=3D 4/3 pi r^3, right? 

No, that's the volume of a sphere. This is a wire. 
(Volume is Pi r^2*length.)

>This gives me mass, right?
>Then mass(9.8 m/s^2)  / 1.0 m = F/l, right? 

Well, basically yes: then I would argue mass*g/length = F_mag/length
(that's just Newton II, the up force balances the down force) and go
from there...  NOTE that although I didn't give you the length (so you
can't find the actual numerical value of mass) it doesn't matter,
because you're going to divide it out anyway, since the magnetic
formula gives you F/length, not F...

CAPA 6, #6

>On number 6, I don't think that I am clear on how the forces behave if they
>are perpendicular to each other.  I can't seem to visualize the B fields
>(overall) in the picture.  They are circular around each wire so I don't see
>how the B field is determined.

The questions are basically all about forces ON the small wire, so the
*only* B field you need to visualize is the one created by the long
wire. And that's the usual old B field (RHR #1, with the formula B =
mu_0 I/(2 pi r)) for a long wire. It runs in circles around the long
wire. Which way is it pointing at various points along the short wire?
It's essential that you figure this out correctly for the rest.  (The
exact value of course varies as you look at different spots on the
short wire, but the *direction* will be the same at all of them)


>B)  IF the short wire moved away from the longer wire, then the net force on
>it would be the same?  This is one where I don't understand how that works.

The formula says that the farther away from a long wire you get, the
*weaker* the B field (due to that wire) is. (RHR #1 tells you the B
field is running circles around the long wire, the formula is just
telling you the magnitude gets weaker, farther away)

Now, if you can picture the B field made by the long wire at various
locations (especially, locations where the small wire is!) then just
think about what the *forces* are on the small wire DUE to that B
field.  Here you use RHR #2, the one that tells you the force on any
current due to an external B field, to give you the direction of the
force.  If you needed to be quantitative, you'd use F = I*l*B, but here
I think direction and rough (relative) size is all that matters.


>E)  IF the short wire rotated around its axis, its rotation would be ccw.
>From RHR's, I say this is true. 

I don't see how you can answer this one until you really understand the
other parts, because you need to think about the forces at various
different points along the short wire to decide if it will rotate or
not. There's no "RHR" that I know of that tells the answer either
way...

CAPA 6, #5

>i'm confused about what this problem is asking and
>what the + - 0 mean. are they effected by another
>force?  i'm a little slow at getting it
>thanks  

The pictures show various particles all travelling through the same,
uniform B field. Some of the particles curve one way, some curve the
other, one of them doesn't curve at all. All the question wants to know
is what is the SIGN of the electric charge on each of the 5 particles?

So if you think particle number 1 must have a positive electric charge,
and #2 is zero (electrically neutral, i.e. it doesn't feel any magnetic
forces at all), and the rest are negatively charged, you would input
"+0---".

Remember, negative charges always curve the "other way" compared to
positive charges in a magnetic field. The quantitative formula (which
you don't really need for this problem) is F=qvB, that tells you the
size of the force. The RHR #2 tells you the DIRECTION of the force (but
it only works right for + charges, it gives you exactly *opposite* the
answer for - charges)

Typo in the notes

>1. On page 20-7 of your notes.. The words at the bottom say the I is going
>into the page - but isn't the symbol telling us it's coming out of the page?

Yes, I got the symbol wrong, the current should've been indicated by an
"X", not a "dot". Sorry!

Curiosity Question

>i was just wondering if there was any physics
>application to the fact (or
>observation) that there is less static on your radio when youre touching
>the antenna. I thought it may have something to do with grounding or
>something like that. Oh, and if there is, is there any way to duplicate
>that touch with something inanimate, (like a piece of metal or
>something) so that i dont have to keep holding my antenna on my alarm
>clock radio to get the static to go away. Thanks, and i love your class!


Thanks for writing! So, first of all, I don't exactly know the answer
to your question, but I can make some educated guesses.  By the way,
after writing this, I went to the "How things work" website (there's a
link to it from our home page, it's a great physics site!) and he has a
lot more *very* readable comments about antennas... Check out  http://rabi.phys.virginia.edu/HTW/  
(select topic :Radio)

FM Radio waves are broadcast in the "100 MHZ" range, give or take, and
lambda*freq=speed=3E8 m/s for electromagnetic radiation. So that makes
the wavelength "meters", about the size of a typical person! So I
suspect that when you touch the antenna, you are simply providing a
bigger and better antenna, improving the reception. (Remember that
although your external resistance is high, inside we're just salty
water, which conducts quite well!)  I think that antennas are quite
sensitive to orientation, and size, so you probably just provide a
slightly better pickup than the small (probably cheap) antenna that
comes with the radio. So I don't think it's "grounding" that helps,
quite the opposite, I think it's the improved antenna.

So.. could you duplicate that with something inanimate? I think so,
buying (or making) a larger antenna, or one that is more highly
adjustable, would probably be quite effective... Even a coat hanger or
other similar sized bendable piece of wire might do the trick.
Commercial antenna's have optimized "impedance matching" to give the
best connection to pickup circuitry, but I've seen people with coat
hangers clipped to their car antennas... The orientation and size you
want depends on the station (wavelength, and the orientation of the
*broadcasting* antennas), so you often have to "fiddle" with the
antenna as you go from station to station too. We're going to be
talking a *little* about this kind of stuff in a couple of chapters!

Talk to you soon - see you in class!

Cheers,
Steve P

CAPA 7, #1

>It is true that an 
>electromagnet with an iron core would be stronger than that with a
>cardboard core?  

Check out the start of the second paragraph of Giancoli section 20-14,
just to "confirm" what you're saying!

>Also, where charges leave would be equivilent to the north end of a
>magnet?

? I don't think there are any *charges* leaving the north end of a
magnet!  (Charges aren't repelled or attracted by B fields, at least,
not if they start off at rest) However, it is certainly true that FIELD
LINES leave the north end of the magnet (on the outside) and head
towards the south end. (See Giancoli Fig 20-3b to confirm that one!)

>If you have a beam of electrons shooting past you, would that generate
>an E-field (if you're not, say, standing IN the beam) 

If I saw a bunch of electrons *sitting still* in front me, Coulomb's
law would certainly unambiguously say that there will be a large
E-field which extends outward to infinity (dying away with distance,
admittedly, but still very much present at finite distances)  So, does
the fact that the electrons are moving suddenly "turn off" this Coulomb
force? Why would it? (If it did, would an electroscope still "register"
while you were moving charges towards it, or only when you stopped
moving? What did you observe in lab 1??)

CAPA 7, # 6 and 7 and 4 together

I am a little confused. On Faraday's law I am confused as to what "N" is in 
>the equation. For example in questions 6 and 7 I don't know what the value
>for N is. Am I supposed to calculate it or is it necessary? I am lost.
>Also, voltage can be a negative value right? 

You're not lost! "N" is "the number of loops", if I don't specify it,
you should always assume it's just "1". So, in both CAPA problems this
week, use N=1. In real life, people like to wrap LOTS of coils to
increase the EMF you get out, so "N" is generally just given to you...

EMF can be negative, but this week both questions asked for the
"magnitude" of the EMF, so that means you should take the absolute
value of your final answer. (Next CAPA you'll get to worry about signs,
or really, direction of induced currents)

>I also had a question about #4, i drew a picture and I thought that I 
>calculated the radius by dividing the distance that they gave us. Can you
>help with that too?

Yes, I agree, the radius in that problem should be half of d.

CAPA 7, # 5

>I have a question about number 5 on this week's CAPA. So the  formula
>is Torque = NIAB.  Torque and I are given.  But I am confused about
>what the radius is.  Would it be 1/2 of the value given?  Also, if the
>loop's face is parallel to the magnetic field, would this make the
>theta 180 or 90?  Thanks-  I think I may just be confused on the
>wording of this problem.

The problem says the loop is "square", so I don't know what you mean by
radius. (The area of a square is easy enough to find, given the length
of a side, no?)

As for whether theta is 180 or 90... I'm going to let you answer that
yourself! Look at Giancoli Fig 20-30. Theta is clearly defined there.
If the "face is parallel to the field", there's no ambiguity about how
the loop is oriented (it's basically Fig 20-30b, right?)  So, what is
theta then?  

Another curiosity question

> >I had a  question that I've been kicking around also:
>>>What is the idea behind polarized plugs (Like your refrigerator cord, or
>>>hair dryer)?What is their benefit, and how do they work?
>>
>> The idea is explained reasonably well in Giancoli 19-9. The third plug
>> in "three pronged plugs" is just a ground, to keep the outside of
>> electrical appliances safe in the event of accidentally internal
>> shorts. For the simpler two-prong polarized plugs, they are designed so
>> that one of them will be "ground" and the other will be "hot",
>> alternating from +120 to -120 every 1/60 of a second. (One of the plug
>> sides is widers, to ensure you plug it in the "right way".) Then the
>> appliance can make sure that outside metal surfaces are all connected
>> to the grounded side, and not the hot side, to avoid accidental
>> shocks.  (The 3 pronged ones are much safer.)

>I  read that blurb in the text, but I guess what I didn't get was how the
>third plug or larger plug is grounded.  I assume it has something to do with
>the wiring of houses, but I remember that hairdryers and such didn't have
>that sort of plug in the past, so it is a somewhat new feature.  The
>outlets at my parent's house, on the other hand, *haven't* changed, so have
>they always had a grounded wire in them, and it just wasn't taken advantage
>of previously?

I should tell you right away that I don't know much about household
wiring.  Anything I say here is mostly guesswork, so don't go wiring up
your house based on it!! 

In most homes, every outlet should have 3 wires coming in to it.
There's the usual (old) two, one of which is "hot", the other is
"cold", and then the third which is just a ground.  (sometimes there,
sometimes not, it'll lead to the "third hole" below the two slots) The
third (ground) is really *ground*, that wire connects, most likely, to
a big copper rod that's been driven into the ground! Or, perhaps, to
water pipes that also go to ground. Or, if you have old wiring in the
house with no true "ground" wires at all, the metal pipes that *house*
the other two wires is probably a decent ground itself, so the new
three-holed plug might just wire the ground slot up to that pipe
itself. Again, I'm not suggesting you do this, proper wiring in a house
should always be done by a professional! But I'm guessing that this is
(logically) possible...

The "cold" outlet is the side with the larger prong, so when you have
plugs that don't have a third "ground" prong, nowadays they will often
still have one slightly larger prong, so you can only plug it in in one
orientation.  The hot side is the slightly smaller hole (so, it's
tougher to reach in there with a little finger?!), and that's the side
that's oscillating up to +120V, and then down to -120V, 60 times a
second. The "cold" side is supposed to stay at 0 V all the time.

But "cold" doesn't necessarily mean ground - remember, there are wires
connecting one plug in your home to the next, and if there's current
running through those wires, there will also be (small) voltage
differences from one plug to the next. So the "cold" side at one spot
in your house may STILL be a few volts away from the "cold" side at
another, and hence is not truly ground. That's why they added the third
prong, those wires should have no current in them, and they go right
directly to ground, so should always be safe to touch.

>Also, what about the outlets that have have the little red and yellow
>buttons you can push, and they usually say something about how important it
>is to test them each month.  What do those do?
>Thanks!

I don't know about "testing"... The buttons that I know about are
"ground fault interruptors". (I think that's the name?). The idea is
that if all is well, the current coming IN one side of your apparatus
must equal the current going OUT the other side. If it isn't, Kirchoff
says current must be going out somewhere else, and that could mean bad
news (it might, e.g., be going through YOU into the floor, ouch!)  So,
the device with the button is measuring current in and out, and if they
don't match, the little button pops out, and all current is shut off to
the device.  (This happens on my hair dryer from time to time, dunno
why. Could be a faulty switch, or maybe there's really a problem and
I'm going to get zapped sometime soon...) When you reset the button, as
far as I know, you just let the current flow again... So like I said,
I'm not sure about "testing" them. 

I'm not at all very sure about this last answer, so if anyone knows
better, let me know and I'll fix this up!

CAPA #8, Question 6

>#6 is also giving me trouble.  The hint you gave said to use
>self-consistent units..I am not clear what that means.  I used the
equation ... (stuff deleted)

"Self consistent units" just means to be sure to use metric everywhere, 
that's all. Your problem is that "mWb" is short for "milli-Wb", does 
that help any?

CAPA *8, Question 1

on #1, I solved it more than enough times, always coming up with the
same, incorrect answer. I solved it symbolically and found that I=... 
(formula snipped)

This problem is not conceptually too difficult, but there are a number
of small pitfalls - it's reasonably hard to pull it all together and
get the right answer!  Work carefully - don't try to "solve this in the
margin", there are enough factors of pi, and 2, etc, that's it's too
easy to slip up!  Solve it symbolically first:  in the case of
this person's email, it was very helpful to me to see a symbolic
formula for the final answer, it lets me check without having to worry
about calculator errors, I never know what numbers you had anyway!


Here are some common pitfalls: Remember to convert diameter to radius
if you're finding areas using Pi r^2.  Convert (carefully!) everything
to metric!  Here's a good one lots of people forget: to find the
resistance "R" of the wire, you need to know the total length "l":
that is the circumference of the (big) circle times "N" (because you
have "N" loops of wire, EACH one has a length = Pi D = circumference)

A tough question

 >    I'm afraid I'm going to ask a why question. 

I'm afraid I may not have a satisfying answer :-(

>Why does a [you need to add "changing" here] flux create
>an EMF?  Is it on an atomic level, for example the electrons being moved
>or a dipole being created in the metal?  I understand why the current
>would create a B-field that fights the change, but it much less
>intuitive why a EMF would be there in the first place to create a
>current. 

First of all, the answer has nothing to do with atoms, because changing
flux in a *vacuum* will create EMF's, without any matter of any kind
being present (that's how EM waves can propagate through empty space).

I have discussed Lenz' law in an "anthropomorphic" sense ("currents
fight the change"), but as you've noticed, you don't always GET a
current, so the EMF doesn't always really fight the change at all. And
of course, my discussion has just been as a tool for you to quickly
recognize the direction of the induced EMF (and/or current); there's
really nothing intrinsically physical about "fighting change..."

I think my (possibly unsatisfying) answer has to be:  at some level,
Maxwell's Equations can't be *derived*, they are merely a concise
summary of experimentally observed behaviour! This is just what
happens, we observe that EMF's are created by changing magnetic flux.
We can write down  a simple equation that describes the behaviour
quantitatively, and then *all* relevant experiments show that it is
absolutely universal behaviour, working in any system, always the same
way!  So that makes us happy - MANY complex behaviour in a huge variety
of systems are ALL explained by one, simple, law.  But WHY that
fundamental law is true.... I don't think I can answer that for you.

I suppose I could argue that physicists have in fact *derived*
Maxwell's equations from some even more basic symmetry principles of
nature (gauge invariance is the buzzword), but I think this won't
really give you any kind of satisfying answer, so I believe the appeal
to experiment has to be my answer:

Let me ask you this: why is F=ma, force proportional to acceleration?
(E.g.  why couldn't doubling the force QUADRUPLE the acceleration?) Or,
why does the electric force drop off like 1/r^2? (Why not like 1/r, or
1/r^3?) Of course, if any of these were true, the world would be a very
different place, and life might never have evolved, we might not be
here to think about it... but still, that doesn't answer the question.
I'm not sure if physicists can answer such "why" questions, when you
ask them at the very *bottom* levels of science. 

I *can* answer *many* why questions for you, we do that all the time in
physics:  Why does an electron curve in a magnetic field? (Because
F=qvB)  Why does the light bulb light up when I hold the antenna in an
antinode in the room with the FM transmitter on? (Because the FM
transmitter makes an E field in the direction of the antenna, and F=qE)
And so on. I answer "why" questions about complicated phenomena by
using just a very few basic *fundamental laws* of physics to explain
them. But I don't know how to explain why those fundamental laws are
true... You have to build from somewhere!

Another curiosity question

 >After talking about light waves and the different kinds, I was curious
>as to how glow in the dark objects work. They appear as a
>yellowish-green, and you stated that yellow means they burn hot, yet
>you are able to touch them.

Nice question, interesting observation, and completely correct!  There
are more mechanisms to get things to glow than just heating them up.
Heating is kind of brute force - it makes all the atoms oscillate
(thermally), and they radiate *many* different frequencies, with a
"peak" intensity at a frequency that increases with temperature. So
e.g.  the sun is hot, and emits radiation at ALL wavelengths, but it
happens to be the brightest in the "yellow" region.

But another way to get atoms to glow is to excite the electrons, and
then let them "fall back down" into their original orbits. We'll be
talking about this soon in lecture.  When this happens, you DON'T get
all those different wavelengths, you only get *one*, very specific,
wavelength. That's what happens with glow-in-the-dark things. (And,
e.g. in my lecture demo of the rock that glowed yellowish under the
black light.  The UV radiation excited the electrons, and when they
fell back down again into the atoms, they emitted only one very specfic
wavelength.)  This process is *not* thermal excitation, and the objects
don't have to be hot at all to do this.

Incandescent light bulbs, by the way, glow because they are HOT, and
because of this are fantastically inefficient. (They throw off lots of
radiation, but only a few percent is in the visible region!)
Fluorescent bulbs, on the other hand, are glowing because high voltage
has excited some electrons, and when they fall back down you see light
from them.  (There are a few different "fall-back" possibilities, so
there are several distinct colors emitted, which together appear
basically white to us) That's why fluorescent bulbs are much cooler,
and more energy efficient. Much less wasted radiation, it's NOT thermal
in origin!

 CAPA 9, #8 
>i just wanted to ask you about this week's CAPA #8.  I guess i'm just not =
>understanding the question, i tried to draw it but when i calculated the =
>angle (how i thought it should be calculated) it came out to be tiny, =
>and i know total internal reflection requires a relatively large =
>critical angle.  is it saying that the light is in the liquid, x cm =
>below the surface, and the ray hits the surface of the liquid some other =
>x cm away 

Yes so far

>and then is all reflected back down into the liquid at the =
>same angle it came in??  

No, when you reach the critical angle, the reflected light reflects, as
always, at the same angle as the incident angle. But you don't need to
think about the relected light. 

Look at figure 23-23. Ray "J" is hitting the surface fairly "near to
straight up", while Ray "K" is hitting farther away.  Focus on ray "K",
the one at the critical angle. That's the ray we're talking about. So
you've been told how DEEP the red dot is, and how far across
(horizontally) the point is where it reaches the top, and you're asked
to find the index.  That's all... (Just don't confuse the angle
theta_c, indicated in that figure, with 90 minus theta_c, the "inside
angle" of the triangle that's been drawn..)




 Question about optional material 
 
>On your web page, you have written that section 23-3 is optional, but today
>in class (Monday) it seemed as though we went through a lot of the
>information presented in 23-3.

Hmmm... No, 23-3 is really optional. 23-3 is all about curved MIRRORS,
and the ray diagrams you need for them. But we haven't talked at all
about mirrors in class (except flat ones awhile ago). We talked
yesterday in class all about ray diagrams for LENSES, which is Giancoli
23-7, 8, and 9.  Some of the pictures in 23-3 may look vaguely similar,
but it's really quite different! (I won't have any *curved* mirrors on
the homework or exam. They're cool though, check out the image formed
by outside car mirrors sometime... Is it real or virtual? How do they
make a mirror "wide-angle" anyway? My little "alligator image" toy I
keep bringing to class is an amazing example of how curved mirrors can
make a real image.)


 Question in class 
I was asked in class which sections we've  SKIPPED  since
Exam II.  Looks to me like we skipped 20.7 and 20.8 (though we talked a
bit about the IDEAS in 20.8), 20.11, 20.15, 21.4, 21.9-15, 22.2, 4, 6,
and 7, 23.3,  23.10 and 23.11.

 CAPA #10, problem #8 
(Someone presented me with a very clear solution to the problem, written
out with only symbols first. Part of my response follows:)

I like your nicely presented solution, it made it easy for me to spot
the error. It's a subtle but important one. You start off, right at the
beginning, with ho/hi = do/di. But that's not right! (Giancoli writes
that down when he's talking about MAGNITUDES of do and di, but then he
immediately fixes up the minus signs)

The correct equation should be ho/hi = - do/di.  And what you're given
is ho/hi = +3.3, which means di = -3.3d0. (you had used +3.3*d0, the
minus sign was missing)  This makes sense, the image in a magnifying
class is *virtual*, so I would EXPECT di to come out negative! If you
now procede with what you were doing, you should get the right
answer... (You're solving two equations in two unknowns, the other
equation being the "lens equation")

By the way, you might notice that the answer you got can't be right...
because you got di bigger than the focal length.  But remember what I
was saying in lecture yesterday, magnifying glasses only "magnify" if
the object is *inside* the focal length!  So I guarantee the answer
better come out *less* than the focal length


 Exam 3 questions 
You said that the test would concentrate  on chapters 20-23.9,
but then you also said that it would cover  everything through
Wednesday's lecture.  So, will concepts from chapter  25 (camera, eye,
magnifying glass) be on the test?  I just want to make  sure I study
everything.

For this material - I'll stick to straightforward "single lens" physics
on this exam.  Camera, eye, and magnifying glass are really nothing
more than *examples* of single lens problems, so there's no harm
reading up on them, but I'll try to stick with the types of questions
you'd expect from purely Ch. 23 material.  (E.g. I *won't* ask anything
specific about cameras like f-stops, or say "angular magnification"
[which is a Ch.25 concept, and differs from "lateral magnification",
which is a Ch.  23 concept!])

Hope this helps -



 Some Exam study questions 
>On page 634 in the book it says Vout/Vin = N out/N in and page 21-12 in
>the notes it's Vin / V out = N in/ N out.  If it's a relationship of
>secondary to primary should the secondary's be on top? I guess
>mathematically it does not matter, right?

Right, it's the same equation, just written "upside down". 

>Should we put the EM spectra on our sheet? The freq and wavelenths of 
>all the different regions?

I certainly won't expect you to memorize any numbers associated with
the regions, but do expect you to have a *sense* of what's long
wavelength and what's shorter, what's EM radiation and what's not...
(And, of course, having a FEW typical numbers in your head might let
you check answers for sensibility...)

>For diverging lens where do < f ---- where is the image? I don't have a 
>pic for this one and I could draw lots of lines...... I will also play with 
>the applet to figure that one out.

I would definitely suggest you play with the 
 applet, or better yet,
"draw lots of lines" till you see what's going on! 

>4. Do we have to know the formulas from 20-9? Torque = NIAB sin theta? 

Yes please. We certainly talked about torque in class, I even derived 20-9a in lecture.

>and M = NIA?
No, we never really talked about the magnetic dipole moment.

Talk to you soon!
Cheers,
Steve P.

 More exam review questions
>1) What sections in chapter 20 are we responsible for on this exam?

We *skipped* 20.7, most of 20.8 (except the basic idea), 20.11, and 20.15
Much of the rest is potential exam material, I guess. 


>2) Which degrees of difficulty do you recommend for the questions at the
>end of the chapter?

I think most of my exam questions so far have tended to be level I and
some level II.  On this exam, I have a *couple* slightly harder
questions than usual. (Mastering level III is generally "A" work, level
II is generally "B" and "C" work, mastering level I should pass you...)

>3) If I know and understand the information on the study guide which you
>put on the web, is there anything else that would be helpful on this
>exam?

Some possible suggestions: best of all: dig up some 2020 exams from
previous semesters (available from my " other links " page, with just a
little hunting), those usually make for nice (and useful) practice!
Giancoli's website also has lots of extra questions, which I sometimes
"borrow" ideas from for the exam.  Make sure you've looked at my
concept questions (on the web, there are lots of extras there too)
Otherwise, I think doing end of chapter problems is also a pretty good
study tool.  



 Set 11, #8 question 
I'm totally frustrated with #8 on this weeks Capa.  I don't even
know how to set up the problem much less solve it with only the
wavelength given.  I have all the other problems done for this week but
am going to tear out all my hair over this one.  I know that m=2 for
the blue wavelength since it is second order but what would m be for it
to be a minimum?  Anyway, I would appreciate a little more guidance on
this problem. 

Well, yes, #8 this week is a real "thinker"!  (But no hear tearing
please, it's only a physics homework problem!) This one is awfully hard
to explain by email, I'd strongly suggest you find a TA to talk to (or
see me after class tomorrow). But let me try to get you started:

You ask "what would m be for a minimum"... I bet you know the answer to
that! Remember, if the path length difference d sin(theta) = m lambda,
when m is a whole integer, then you get a maximum. So... what is the
condition for a minimum? (What does the path length difference have to
be to get *destructive* interference?) Have you looked at Eq's 24-2 in
the book?

One thing that might help next is to *sketch* a picture of what the
screen intensity looks like, first just for the blue light. You could
label the location of the 2nd maximum - call it "x" or something. You
don't have a number, just give it a symbol.  I might suggest writing
down the *formula* for "x" in terms of the usual things (L, lambda, and
d of the slits) It'll be fine to use the "small angle"
approximation...  Many of those are NOT given, but that's o.k, just
give them names.

Now sketch a picture of what the screen looks like for the other color
light.  There are *lots* of possibilities (an infinite number) that
have a minimum at that spot you labelled "x". Draw a few of them! ....
Pick one, and try to derive an analagous formula to what you had
before...  i.e. a formula for "x" in terms of the usual things, this
time realizing that x is supposed to be the *minimum*.  But "x" is the
same in both pictures, and so is "L" and "d", so perhaps you can set
the "x"'s equal in your two formulas, and all the many unknowns (like
"L" and "d") will CANCEL, leaving you a formula for the new lambda...

Like I said, there will be many possible lambdas (one for each
different picture you drew that had a minimum at that point "x"), but
only one will be "visible light". The rest will either be too *large* a
lambda, or too small, but not in the "visible" range...

This is pretty hard to explain by mail, a blackboard helps a lot . Hope
you can get started!  (By the way, this is basically problem 10 from
the end of the chapter)


 CAPA #11, Problem 3 
I am having some problems with CAPA set 11 #3. I am using the
equation m=fo/fe (my numbers are .874 m/.0286 m = 29.62) and CAPA is
not taking my solution. Is the equation wrong or is my math messed up? 

Check out Giancoli Eq. 25-3, your equation has one important small
mistake...

 Grading questions 

An anonymous question:I have a question regarding CAPA. I know that
>you drop the lowest score, >but what if we get 100% on all of them do
>we get extra credit?

Hmmm. No, I'm afraid not. Getting 100% on all of them is a nice
achievement, though - hopefully it means you're learning so much
physics that you're acing the exams, and so don't NEED any extra credit
:-) If you're getting 100% but are not doing well on the exams, you
need to think about what's going wrong on the exams, we should probably
talk in person.  (If you've been getting 100% by virtue of getting some
help, I suggest you try doing the last few CAPA sets more on your own.
If that lowers your score a little - no worries, because I will drop
your lowest score, so you can afford to take a little risk at this
point!)




 Question # 4 : CAPA BUG - PLEASE READ!  
 Someone emailed me with a nice detailed (and correct!) explanation of
what they did, but CAPA said no 

This was a bug in CAPA! It will only affect a very small number of
people, but please read this if you were getting an "N" on this problem
BEFORE Wednesday evening:

CAPA had been programmed with a value of h*c which was off by only
0.2%, but in your case because hf and W were so close, this tiny offset
in hf made a (relatively) large difference in the final answer, namely
CAPA was off by 2.4%. Since it requires YOU to agree with IT at the 2%
level, it said you were wrong! (This happened because, by random luck
of the numbers, in your case you were subtracting two *very* similar
numbers to get a much smaller final answer)

I have fixed the bug, so just try re-inputting your old answer, it
should say yes now..

PLEASE NOTE: This bug was fixed Wed at 6:35 PM. If you enter an answer
AFTER this, and it says no, you're still  really  getting it
wrong!!  If you entered your answer BEFORE this, well... you might have
been right or not. Better try again! Do watch units here, and keep
plenty of sig-figs. Let me know if you have any questions/problems with
this one (but only if you started entering answers before Wednesday!)


 A "philosophy" question 

> I was thinking about what we were speaking about in class and I
>was suddenly stuck.  when we think about a wave with no mass, what
>exactly are we talking about.  Is it just vibration and energy or what
>exactly?  Maybe there is no exactly which I'm willing to accept as
>well.   I suppose that this is why the distinction between matter and
>energy becomes a bit blurry?  Can you think of matter as being  slowed
>down energy in a sense?   

I don't completely understand the question, but let me try .... a wave
without mass... I don't know how to answer "what are we talking about
exactly", because classically waves have nothing to DO with mass.  In
our usual mental models, you either have a wave (which is not a
particle, so it has no mass, that's not a PROPERTY you usually
associate with a classical wave) OR you have a particle (which usually
has the property of non-zero mass). And now you have light, which is
both particle and wave, and when you consider it a particle, the fact
that it travels at c means it cannot have any mass... I don't believe
you can think of matter as "slowed down energy", because matter can
also HAVE energy (e.g. kinetic or potential) which we think of as a
property of the matter, rather than as being the matter itself...

The real answer, I suppose, is that I'm trying to use  words  to
explain what is in reality a detailed mathematical theory (Quantum
Electrodynamics, or "QED") which allows you to compute the behaviour
and interactions of light and matter. And, I'm at a bit of a loss to
try to provide words that connect this behaviour to everyday
(macroscopic) phenomena.  But I think, in the end, that one needn't
philosophize too much about all this weird stuff. We have created a
mathematical/physical framework (QED) in which to calculate, and then
you can make quantitative predictions about experiments: that's pretty
much what physics can do for you. The fact that we can't "picture" or
"understand" what exactly we're talking about is in a sense irrelevant,
as long as we can make complete and accurate predictions about
experiments and devices... This is precisely what some philosophers
DON'T like about quantum mechanics - we've partly given up trying to
picture what's "really going on", but are still able to predict things
very accurately - it bugs a lot of people!

Hmmm...  help any?  Try to reformulate your question, and I'll see if I
can provide a more satisfying answer!  (Of course, if you switch to a
physics major, go on to grad school, I'd say in about 6 years you'll
have a very thorough understanding of photons, at least mathematically!)


 


 CAPA #14, problem 8 
>#8 - I don't think I understand the question.  I have the radius of a
>neutron.  P = mv.  I have m, but I can't figure out how to get v.  I am
>stuck on this one....

You have the nuclear radius (not the neutron radius. A nucleus is much
LARGER than a single neutron) The idea is that you know a neutron is
SOMEWHERE in the nucleus, but you're never sure where. (It's a quantum
wave, it's SOMEWHERE in a box the size of the nucleus, but effectively
"smeared out" like any quantum particle)

So there is a "delta x", which IS the nuclear radius, that's the
uncertainty in where the neutron is... Given delta x, the Heisenberg
Uncertainty Principle (HUP) gives you an uncertainty in momentum.  My
hint was that if you don't know which way something is going (the
neutron is bobbing around in the nucleus, you don't know where, or
which way), it's quite reasonable to argue that a typical VALUE of its
momentum will be numerically similar to the size of the UNCERTAINTY in
its momentum. (i.e. p = delta p, or equivalently, there is 100%
uncertainty in the momentum) 

Since HUP gives you the uncertainty, setting p=delta p gives you p, and
then it's easy to compute the energy. (You have to look up m for a
neutron) Once you get this problem, think about how remarkable it is.
The HUP is NOT just some vague statement of how poorly we physicists
understand things. You just *used* it to compute a measurable quantity,
the energy of neutrons emitted from radioactive nuclei! The calculation
works, your answer is very close to the experimental result.

>One more capa question and I am done forever.

What, you mean you're not planning on taking Phys 2130, intro to modern
physics for non-physics majors :-(?  

Here's a plug: how about the course I'm teaching next fall, Phys 3070
(Energy in a Technological Society) a totally *non*-physics major
course.  That one *won't* be using CAPA for homeworks!



 Final Exam Question 
>Would you mind telling me what sections from the new information is on the
>final.  Thanks

New material since the last exam would be:


Ch. 25 We covered 1-4 (eyes, cameras, magnifying glasses, telescopes),
I won't ask very tough questions about this because we kept that
discussion (and CAPA problems) pretty basic and mostly qualitative too,
but some questions are possible (e.g , something very basic about
"nearsightedness", or camera focus, or angular magnification, or
something like that.  You did a lab with a simple telescope too, such
material would be "fair game"!)

I never assigned 25.7 and 8 as reading (angular resolution of eyes and
telescopes), but we did have a discussion of it in class, and in the
notes (and a concept question or two) and also in lab 5. So again, I
wouldn't expect anything *too* detailed, but certainly could ask about
the *ideas* there.

Then we went back to Ch. 24 (wave nature), we skipped 24.9. For 24.8 we
really didn't do any quantitative CAPA problems, but did spend some
lecture time on it (so I wouldn't ask any more than a basic qualitative
question about "soap bubbles" or "antireflective coating" etc. )

We skipped CH. 26 (We have used the formulas "E=mc^2" for a particle at
rest,  and "E=pc" for photons, from that chapter)

We did everything in Ch. 27 except 27-7. 

We'll end up covering 28.1-7.  You can tell from the last homework
where the emphasis on that quantum stuff will likely be (Mostly
Heisenberg uncertainty?). We haven't spent as much time on Ch. 28 as on
the others, so it won't be as prominant on the final.

The final will be more than half "new material". The rest will cover
everything from the whole semester.



 SO many questions!! 

>
>The relationship is d sin theta = m lambda
>        1. When lambda increases, d stays the same --- theta increases and
>interference spreads out, correct?      B/c sin theta = m lambda/d and if
>lambda increases and d doesn't' change then the angle would have to     get
>larger.

Yes. E.g. Red light on a given grating is spread out MORE than blue light.

>
>        2. If you have a big slit you have a smaller lambda, correct? Then

Lambda is up to you, it's whatever color you choose to shine on the slit!
You can shine ANY lambda on any slit..

>you
>won't see a diffraction pattern --      b/c the pattern will be too small?

But yes, if you choose to shine a small lambda on a big slit, the light
behaves like you'd expect geometrically, the wave nature is not
apparent. (The diffraction lines are so closely spaced they merge
together) You see a normal "shadow" of the slit...

>
>        3. If lambda is larger than the slit then you only see a broad
>pattern?

Yes.

d sin(theta) = m lambda, so if lambda is large, theta gets large. If
lambda is larger than d, then you can't even solve this equation for
any theta (because sin(theta) is never larger than 1!) so your pattern
has become so broad you don't even get to the first maximum...

>
>        4. When you get to electrons and the like - detection of interference
>only
>occurs in the slit distance is  less than or equal to  lambda, correct?
                    

It's not quite so formal as all this. You can see interference if you
look carefully enough even when lambda is much smaller than d, but it
just gets EASIER and more noticeable to see interference as lambda get
bigger. The "generic scale" where it's quite visible without any fancy
tricks is when lambda is about the same order of magnitude as the slit
size...


>Correspondingly in order to detect an electron (etc) the speeds         would
>have
>to be very slow in order to achieve a detectable lambda, correct?

? You can always detect an electron, independent of the speed. If you
want to see electrons INTERFERING with one another (i.e if you want to
observe the "wave nature" of electrons), e.g. giving you some
bizarre "minima" where no electrons go, then you want lambda to be
comparable to the slit spacing, which generally means pretty slow. (Not
THAT slow though, because electrons are very light)

>
>        5. Single slit grating:
>        * slit gets big - pattern looks like a shadow of a slit         (how
>        * big        

No definite answer to "how big", generically, when lambda is "much
less" than the slit opening.

>* Slit gets small - very wide pattern
>        * (how small is small?)

As usual, no definite answer, but small means "comparable to lambda".

>        * fixed slit, longer lambda to shorter lambda - pattern squeezes in
>        (why?)

First minimum (which in a real sense defines "how broad the pattern looks" is
Dsin(theta) = lambda,  or if you prefer, D (x_1st min /L) = lambda

So, if lambda is bigger, x is bigger, i.e. the pattern is broader. 
As lambda gets smaller, x gets smaller, the pattern "squeezes in".  

That's the mathematical answer. The physical answer is that the SMALLER
the slit, the more POINTLIKE the source, and pointlike sources send
waves out in ALL directions including very large angles.  (Huygens says
a point source makes a perfectly spherical wave)

Bottom line: small slits diffract light MORE strongly.

>        * fixed slit, shorter lambda to longer lambda - pattern spreads out
>        (why?)

Same as above...

>
>2. Are you going to cover telescopes on the final? In any event, the angular
>magnification for a telescope is M = theta prime/theta = - Fo/Fe.....The M =
>N/f only applies to magnifying glasses right?

Yes, I will, and yes, that's right.

>
>3. Lecture dated 4/12 - At the very beginning of class you drew the double
>slit diffraction grating.  You were talking about the CD and a prism and
>their differences.  You said they work the opposite way.  Can you explain
>that.  I understand the prism pretty well - it's the CD....  You also said
>if m = 0 then there is no separation of color....Can you explain.

Diffraction gratings of any kind (slits, CD's, whatever) spread out red
light MORE, to larger angles. It's from the formula 
d sin(theta) = m lambda, 
so if lambda is bigger (redder), theta is bigger (more deviation, 
more spreading)

Prisms tend to shift the direction of red a little, and the blue MORE.
That's an *accident* of nature, the fact the index of refraction in
many (not all) materials is slightly larger for shorter wavelengths.
(So blue sees a slightly *larger* index than red does, which tends to
refract the blue light MORE strongly)

If m=0, dsin(theta)=0*lambda =0. So ANY wavelength light will travel
through in a straight line (theta=0 is the solution to that equation!)
So white light doesn't get clor separated, ALL the wavelengths travel
exactly the same into the m=0 line...


>
>4. Double Slit Grating:
>I have in my in notes 1/2x = d sin theta = 1/2 lambda - I don't understand
>this - would this be the point of a dark spot?......Wouldn't x be (m *
>lambda * L)/d. And we can use this to when the angle is less than 10
>degrees, correct?  If the angle is larger than 10 degrees we go back to
>using sin, correct?

I'm not sure I know what that formula is either, I don't understand the
"1/2x" at the left. The formula d sin(theta)= 1/2 lambda would be the
condition for a minimum (dark spot) for a double slit.  (Perhaps you
were making notes for single slit? Even so, I'm not 100% sure what that
formula is referring to)

Around 10 degrees is indeed the cutoff for using the small angle
approximation, (sin(theta)=tan(theta)=theta = x/L, in which case 
x = m lambda *L/d for bright spots in a double or multiple slit example.)


>
>5. Can you define diffraction limit.

When light goes through any slit, it diffracts. The "diffraction limit"
for a telescope refers to the SMALLEST angle you can resolve, that
means the smallest angular separation two point objects can have and
you can still TELL there are two objects. If they get closer, to LESS
than your diffraction limit, then the diffractive "smearing" of the two
images make them overlap into one big fuzzy blob, and you can't tell
there are two things there.

The (rough) formula is angular separation must be greater than
lambda/D, where D is the diameter of the telescope.

>
>6. Page 24-19 of your notes - You don't talk about the relationship of n2 to
>n1 for say oil on water - the book on page 741 goes into detail about this -
>do we have to be able to do this?

I probably won't be asking any quantitative questions about "thin film"
interference, if anything just something qualitative relating to the
"bubble" demo or something. I don't think I assigned any CAPA questions
on thin films..

>
>7. Polarization: Say I have three polarizers - after the first polarizer I
>have Io/2, then the second polarizer is at 60 degrees so I have (Io/2) cos^2
>60, then the third polarizer is at 20 degrees so finally I have (Io/2 cos^2
>60) cos^2 20.....I guess the question is - as you progress through
>polarizers what you add to the next equation is the cos^2 angle of the new
>polarizer, correct?

Yes, cos^2(angle of the new polarizar WITH RESPECT TO THE PREVIOUS
ONE), the angle is always the angle between "direction of polarization
of incoming light" and "direction of new polarizer". (So, theta is NOT
the angle with respect to, say, some fixed direction...) I think you
understood this though. So in your case, if the first polarizer was
vertical, the second was 60 degrees from vertical, the third one must
have been 20 degrees FROM THAT LAST ONE (i.e. either 40 degrees, or 80
degrees, from vertical, I can't tell from what you told me). Make
sense?

>
>8. Page 27-11 of your notes.  The whole point is that momentum of photons is
>quantized, correct?  I guess also conservation of energy applies and the
>only thing that changes is the frequency  of the photon and consequently the
>lambda of the photon, correct?  This is non-classical b/c classical says the
>frequency wouldn't change...ie wave in wave out....Right.  This section
>isn't coming together for me.....Also the last statement on that page -
>conservation of momentum - you can predict angles - the angels generated by
>the new frequency?    So if you used P = mv instead of using P - E/c then
>you would not be able to predict the angles, correct?

Momentum is "quantized" in the sense that p = h/lambda is fixed (given
the color, i.e. given lambda).  When light scatters from an electron,
it gives energy to the electron (the electron recoils with KE, after
all), so the light must LOSE energy. But light cannot slow down, E =
hf, so if the photon loses energy, its frequency must go down. That's
conservation of energy. Yes, classical says frequency doesn't change,
color in = color out...

Conservation of momentum says p_init = p_final. p is a vector, it has x
and y components. Each is *separately* conserved. So if you measure the
energy and angle of, say, the recoiling electron, you know exactly its
x and y momenta components, and then the energy and ANGLE of the
recoiling photon would be completely determined by conservation of
energy, and p_x, and p_y. I wouldn't ask any quantitative questions
about this, but would expect you to understand qualitative aspects
like: the more energy the electron gets, the longer the wavelength of
the final photon, stuff like that.

(If you tried to use *non*relativistic kinematics for the photon, none
of the details of the prediction of angle above would work)


>
>In my notes before the photon crashes into the electron ( I have x-rays
>coming in and longer waves coming out) I have E(photon) and P(photon) coming
>in --- Is this supposed to represent E * P? Or just that they are both
>coming in with the photon?  I think they are coming in separately.

They are separate, different things. A photon has E=hf, and
p=h/lambda.  Two separate properties, just like any particle has
energy, and momenta. They are separate, but *related*, by E = |p c|.
(For ordinary nonrelativistic particles, the connection would be E =
p^2/2m)

>
>9. Relativistic:  light c = v
>E = hf for one photon
>P = E/c = h/lambda
>
>Non-relativistic (v< c)
>E = 1/2 mv^2
>lambda = h/P = h/mv
>
>But for relativistic lambda also = h/mv....
 
No, it doesn't! A photon has m=0, and v=c. You can't use lambda=h/mv for
photons.

>So the difference is that for
>relativistic c = v and for non-relativistic you need to determine v which
>should be less than c....Right? 

No, we really haven't covered what to do if v is VERY close to c. Just use
the formulas E=hf and P=E/c = h/lambda for photons and you'll be fine.

>How much less?  Is non- relativistic v = 0 m/s to but not 
>including 3e^8????

No sharp cutoff, but generally any speed less than 1% of c can be
safely considered "non-relativistic"

>
>10. On your notes page 27-18 - the predicted size of atoms is 1/2 angstrom?

Yes, I blew it TWICE in my notes, claiming the size of atoms was 5E-10,
the correct number is .5E-10

>
>Thanks

Phew! Hope this helps!
Cheers,
Steve P.

 More questions, this time on quantum stuff 

>
>Yesterday in class when we started talking about Bohr and r * p = Hbar *
>n - You said if you speed up, you double ________???

I'm not sure what I said. Here's the bottom line: if you go to higher n
in the Bohr model,  it's complicated!  Radius increases (r = n^2 * 0.5
Angstroms), energy increases, but in a funny way, it becomes LESS
negative. Velocity you can get from r*p = n hbar (that's the
quantization of angular momentum formula), so m v = n hbar / r, and
then you have to plug in for r = n^2*0.5, to get m v = hbar / n * 0.5
Angstroms (notice the n in the numerator cancels one n in the
denominator, but there's one left behind). So if n increases, velocity
actually DECREASES. (Far away orbits, the electron is moving more
slowly)


>
>Your notes Page 28-04 - Do you mean that you can't locate things if
>lambda >> objects size b/c long wavelengths mean bad resolution? If not,
>then what?

Yes, it's hard to look at small objects using long wavelengths.  Long
wavelengths have worse resolution is a fine way to put it. Generally, if
lambda is much bigger than the size of the object, the wave doesn't even
NOTICE the object! Visible light is useless for looking at things smaller
than a few thousand angstroms. To look at ATOMS, e.g, you need high energy
x-rays (or else, an electron microscope with lambda_de Broglie small) 


>
> What is the quantum limit - I mean I understand what it is -but is there
>a value (numerical) to it?

No value. It depends on the situation. Generally, atomic sized things
are "quantum". (But e.g. Prof Wieman over in the Jila tower has managed
to create a macroscopic (mm sized?) chunk of "quantum matter", a
Bose-Einstein condensate, in his lab!)  Generally, when the product of
sizes and momenta get close to the HUP limit, x*p around hbar, life
gets "quantum-y".




 Final exam (generic) 
>Is there going to be a long answer question on the final?
No, the final will be (36) multiple choice questions.

>What's the story on the crib sheet for the final?
Two pages (handwritten) are allowed this time

>I was just wondering how specific the information on the old
>stuff will be on the final.

I'm not sure what you mean. I'll have both number crunching and
conceptual questions on old material, as usual, covering back to the 
start of the semester. It'll amount for about (a little less than) half 
the final. Anything you've seen on old exams would be fair game, but
I'll of course have other questions (similar in difficulty, just
different) too. In terms of "how specific", well, I would certainly 
expect you to understand and know how to use any formulas from my "old 
crib sheets" from throughout the semester.  

 NEW: A question about diffraction 
>What is the difference between diffraction patterns and interference 
>patterns?  

I think of them as closely related. Technically, diffraction refers to
the spreading out of light when it encounters a small opening (or sharp
edge). Interference refers to the resulting bright/dark pattern that
arises when two different sources come together. But then we use the
word "diffraction pattern" to refer to the "inteference pattern" caused
when light diffracts through small slits!

Here's a case where you could have interference without diffraction: if
you take two beams of light, split them with a prism or "half-coated
mirror" (that's not diffraction!), run them through different path
lengths, and then recombine them , you would see an interference
pattern (two sources of light, out of "synch", making light and dark
spots) but it's NOT caused by diffraction.  (There's no "spreading out
from a slit" in this example)

>I have in my notes that you can only have diffraction 
>patterns with monochromatic light...if so, when do you have interference
>patterns, and when can you have both?

No, diffraction does  not *require* monochromatic light, it's just a
whole LOT easier to see/notice if you have only one lambda. That's
because different wavelengths diffract a different amount, so if you
have light that's NOT monochromatic, any interesting diffraction
patterns tend to get "smeared out" and hard to notice. But remember,
one day in class I shined a white "mini-Mag light" flashlight at a very
good diffraction grating.  What we saw was a white spot in the middle,
and then faint rainbows off to the  side. So there *was* diffraction
and interference going on with white light! (Also, a CD will make nice
diffraction patterns from white light sources too) Diffraction means
the white light got spread out by the grating, interference means the
light coming from *different* spots on the grating recombined, "adding
up" or else "cancelling out" at various points on the blackboard. 



 NEW:  A question from someone reviewing interference 
>>    When we are using the equation dsin*theta=m*lambda, is theta in
>rad's or degrees or does it matter?

In this form, it doesn't matter. If you want to use the small angle
approximate sin(theta) = theta, (or equivalently, theta=x/L) then theta
*must* be in rads...

>   I was reviewing the capa over this material and on set #12 prob #4
>you use the minimum 5 ticks away in order to find the width of the
>slit.  Why is it not the first valley?

I'm not 100% sure what you mean, but I'll assume that by "first valley"
you mean the VERY first minimum (the one occuring at approximately 1/2
a tickmark?) That valley is *not* arising because of the width of the
slit.  Those narrow, closely spaced peaks are all arising simply
because you have TWO slits. Those frequent maxima are located wherever

d sin(theta) = m lambda

and the "valley's" you referred to are located whereever
d sin(theta)=(m+1/2) lambda.

Since "d" is relatively large, that means theta will be correspondingly
small (and closely spaced), you see lots of peaks, closely spaced, all
because of "d"...

So, how does the width ("D") of each individual slit enter? It's what
causes the "envelope", the gradual decrease in the height of the bright
spots (and then farther away, the "rebirth" of them) The envelope in
problem 3 and 4 is, in fact EXACTLY the curve given in problem 2, which
is just the shape caused by a single slit by itself. Bottom line: if
you want to learn about the SINGLE slit shape, you have to look at the
envelope. That's why I went over 5 ticks, to find the first minimum of
the *envelope*. (If you just want to learn about the DOUBLE (or
multiple) slits, then you look at the more closely spaced peaks and
valleys.)

It's an important point, so it's good you asked about it. Let me know
if you're still not getting this! Giancoli doesn't talk about this
problem much, but Lab 5 (on page 5.4) has a very nice picture and
explanation.

 NEW: more questions 
>1)  There are some things not on the crib sheets which have been
>assigned as readings.  For example - the information on mass
>spectroscopy and radio and television waves.
>There have also been things not assigned in readings which have appeared
>on the exams - like the information on the hall effect.  (If I am
>incorrect about this part, I apologize).

Well, on crib sheet for Exam 3, the last two entries for Ch 20 were
"Crossed E and B (velocity selectors), v = E/B. 
In uniform B: R = mv / q B.  (circular path)"
Which is really all you need for mass spectrometers....

Radio and TV waves were "cultural readings", I thought you'd appreciate
having some idea how they worked, but won't be specifically testing you
about those *details* on the final. (I assume you're referring to the
stuff in 22-8?)

I'm not sure what you mean about the Hall effect appearing on exams -
which problem are you thinking about? (I can't remember talking about
the Hall effect in class, or testing on it?)

>2) The negative sign in Faraday's Law
>It reminds us that the induced current is in the opposite direction of
>the initial change in flux.  How should it be used when numbers are
>inserted?  You kept saying to ignore it and just know what it stands
>for, but then how do you know when the answer is negative or
>positive???????

When I've asked Faraday's law questions, I've tried to be careful to
always ask for the *magnitude* of the induced voltage (or induced
current), so that you wouldn't have to worry about the sign! When I
want to know about the sign, that's when I ask "is the induced current
clockwise or counterclockwise" type questions!

>3)  If I and B are known, can the direction of E be determined???

Hmmm... I'm not entirely sure what you mean. Maxwell's equations
ultimately answer any such question, but as far as Phys 2020 is
concerned, if I ask questions about "E" it'll probably be in the
context of CH. 16 or 17, i.e.  figuring out E from Coulomb's law (from,
say, a couple of pointlike charges strategically placed!) or in a
capacitor. We really haven't solved for E in any harder situations...

>
>4) Could you please explain Maxwell #1 and #2 in a little more detail.
>The simplicity of the statements doesn't really mean anything for me
>(i.e. a special case of Coulomb's Law)  I know what Coulombs Law is, and
>the equation for it but don't understand what it means with regards to
>Maxwell.

I'm happy to explain Maxwell #1 and #2 in more detail, but it might be
a little tough to do by email. As far as the *final* is concerned, for
Maxwell #1 and 2 you *just* need to know Coulomb's law, and how to use
it.  

But if you're curious, Maxwell's 1st law is explained in Appendix D
(p.  1055). It goes well beyond what we've covered in class. Gauss' law
is just another name for Maxwell Eqn #1.  Remember, Maxwell didn't
invent any of "his equations" except PART of the 4th equation, but he
pulled them all together and understood the connections between them
(and consequences of them) that had never been appreciated before. So,
Gauss' law is a *generalization* of Coulomb's law: it tells you about
the electric field around any arbitrary distribution of charges (not
just a *simple* collection of pointlike charges, which is what
Coulomb's law is good for) Maxwell #2 we hardly talked about. At its
simplest, it states that there are no magnetic "charges", or monopoles,
in the world. But it too can be generalized to help you calculate
magnetic fields. Anyway, this is probably best for "summer office hour
visits if you're still curious!"

>
>5)  How much about the eye do we need to know.  The book goes into
>detail about diseases and the like but that wasn't ever really covered
>in class or on the crib sheet.

The eye is basically just a simple converging lens, so it makes a nice
*example* for CH. 23 ideas. You should probably understand the
*concepts* involved with the two major "diseases" (if you can call them
that), myopia and hyperopia. You don't need to memorize *any* words,
just look at the pictures in Giancoli 25-2 and understand the  physics
involved! (The physics has nothing to do with the eye, per se, it's
really just basic properties of lenses...)

>Last......
>6) If light has been polarized, why can it be polarized again at a
>different angle?  Does the intensity decrease?  I guess I don't
>understand how if all the light except that with a wave in one
>orientation is removed, light with a wave in a different orientation can
>then be removed.
>

It's a good question! First, yes, the intensity decreases (Iout =
In*cos^2(theta), and cos is always less than one, so Iout is always
less than Iin). The reason that you can get light out at another angle
is that light is really an electric field *vector*, and you can always
take the COMPONENT of a vector in any other direction!  The component
will always be smaller than the vector you started with. I suppose it's
like you're taking a *piece" of the E-field, but I agree it's a bit
hard to picture. If the E field points up, and you run it through a
polarizing filter that's tipped at 45 degrees, you let through the
COMPONENT of the E field vector which is parallel to the filter, and
you "chop out" (absorb) the component that's perpendicular.  I'm not
sure if I'm giving you a satisfactory answer at a gut level. Let me
think a bit about whether I can come up with some analogy to make this
easier to picture.

>-also- why, when only one polaroid is used does Iout = .5Iin?  Shouldn't
>it be a whole lot less because a majority of the orientations of the
>wavelengths have been removed???

Again, looks like you're thinking of a polaroid filter as removing ALL
light except light which is PERFECTLY aligned, but that's not what it
does.  It only removes the *components* of light that are
perpendicular. It leaves behind plenty of light, all the parallel
component gets through. Look at Giancoli Fig 24-39.  The light (E0)
coming in is tipped, goes through a vertical polaroid. It's not
entirely chopped, the COMPONENT of E0 parallel to that filter makes it
through just fine.  So the majority of orientations aren't fully
REMOVED, but you only get their "vertical" component (if you filter is
vertical).  On average, half of a random vector's component is "Up
down", the other half is "left right", so if you run random light
(unpolarized) through a filter, on average half makes it through. The
cool think is that whatever makes it through is now 100% polarized!

Polarizing filters are something many students don't think much about,
just memorize the one formula. But they're quite remarkable, and as
you've noticed, pretty counterintuitive. It's very useful to think
through what's really going on, like you are, in the end it makes that
formula make a lot more *sense*!

Hope I've answered your questions! If you think I missed the point on
any, let me know and we'll try again.

Cheers, Steve P.


 NEW: Bubbles 
> I noticed that you mentioned the likelihood of having concepts about the =
>bubble demo and antireflective coatings for lenses on the final in =
>response to other questions.  Looking in the book, these were covered in =
>24-8, a section I thought we did not cover.  So, how much should we know =
>about these topics?  This section talks about if n2 is greater an n1 the =
>ray changes phase by 1/2 lambda-is this relevant?  And about the bubble, =
>why was the top of it black, and the spectrum only at the bottom?  I =
>just don't really understand these two concepts so any explanations =
>would be greatly appreciated! 

I didn't assign 24-8, so I won't be asking any *detailed* questions
about bubbles, or that "phase flip" business. It was a cultural
benefit lecture, so you knew what was going on with colorful bubbles!

However, having said that, realize that (other than the subtle detail
of the phase flip at reflection) the *idea* behind antireflective
coating is a straightforward application of the main concept of CH 24,
that light can destructively interfere. The idea that if light
travels two different paths, and the path length *difference* is half a
wavelength, then the two light beams will cancel out, that's a central
idea.

So, you won't need this for the exam, because of the phase flip story
that we really didn't talk enough about. But as for why the bubble is
black at the top... It's VERY thin there. So thin, that t is much less
than even one wavelength of light. Light can reflect off the outer
(1st) boundary and head back, or it can make it into the soap, and then
(a tiny distance later) reflect off the BACK side, and head back.  So
there are *two* beams that can reflect back towards you. They have
almost ZERO path length difference (because 2t, the path length, is so
tiny). You'd think that means constructive interference, a bright white
reflection. But instead it's black, NO light reflects! (Actually, it
reflects but "cancels out" by destructive interference!)  That's where
that subtle business of the "phase flip" comes in. One of the two beams
(the one the reflects off the outer boundary, where n2>n1) gets a 180
degree phase change (Giancoli Fig 24-31), i.e. it flips sign when it
reflects. The other beam does NOT, because it's bouncing off a boundary
where the next n is smaller. The two beams are exactly out of sync.

Anyway, if you're curious, I'll explain it better in person, it's
really not that hard, but we just didn't talk about it much in class!
(A picture helps a lot on this one)

 NEW: More review questions 
>The exam problems that I was thinking about are 1-4 in version 1, exam
>3.  Both protons and electrons were used.  I guess I'm missing some
>fundamental difference between the problems and section 20-11.  Could
>you explain it to me?

Those questions don't have much of anything to do with the Hall effect
(although the pictures admittedly look similar!). The Hall effect has
to do with flow of current in a conductor (like through a wire).
Questions 1-4 on exam 3 were more directly related to the discussion at
the start of 20-12.  It's just about "crossed E and B fields".

The essential idea is that if E and B are crossed (perpendicular), and
you send in particles perpendicular to BOTH fields, you can tune E
and/or B so that the particles go straight through. For this to work,
you need the electric force FE (=qE) to exactly cancel the magnetic
force FB (= qvB).  (You have to use the right hand rule to see the
directions involved, that's what exam question 1 was about!) 

The formula says qE = qvB if the forces balance, notice that the charge
cancels, so E = vB.  (That was exam question #2)  This tells you that
it DOESN'T MATTER what the charge is, it cancelled out. ANY particle
(any mass, any charge!) will go straight through provided only that it
has the right SPEED, v = E/B. (that's what question 4 was about).

Does it make sense to you? Take a look at those problems again, see if
you can work them out and understand them all. Please let me know if
I'm still not explaining it to your satisfaction!


 A review question 

>On the first exam - Question 14.  (001) we are looking at the
>parallel plate capacitor - you put an alpha particle at the top
>plate....It we do work on the particle to get it to go to the negative
>plate it will go - but why would it accelerate? Wouldn't it want to be
>at the positive plate.  I got the answer right - but just trying to get
>the concept.

You don't need to do work on it, it will *spontaneously* go.

An alpha particle has charge q=2e (which is POSITIVE, remember! An
electron has charge -e) So it naturally wants to go from the top
(positive) plate to the bottom (negative) plate. You can think of it as
being repelled by the + plate and attracted to the - plate. Or, you can
think of the uniform E field throughout the capacitor, pointing down,
which makes F=qE = (2e)E (down). Either way, there is a net electric
force DOWN on the alpha particle, which causes it (by Newton II) to
accelerate all the way, gaining kinetic energy all the while. (It loses
potential energy, and in the end total energy is still exactly
conserved, of course)

I'm wondering if maybe you just thought the alpha was
negatively charged? It's not. If it *was* negative, it would just sit
there at the top plate, and not budge, and you couldn't answer the
question without lots more information, like the exact external force
being used to drag it away from that plate 



 Final Exam Question 
>Would you mind telling me what sections from the new information is on the
>final.  Thanks

New material since the last exam would be:


Ch. 25 We covered 1-4 (eyes, cameras, magnifying glasses, telescopes),
I won't ask very tough questions about this because we kept that
discussion (and CAPA problems) pretty basic and mostly qualitative too,
but some questions are possible (e.g , something very basic about
"nearsightedness", or camera focus, or angular magnification, or
something like that.  You did a lab with a simple telescope too, such
material would be "fair game"!)

I never assigned 25.7 and 8 as reading (angular resolution of eyes and
telescopes), but we did have a discussion of it in class, and in the
notes (and a concept question or two) and also in lab 5. So again, I
wouldn't expect anything *too* detailed, but certainly could ask about
the *ideas* there.

Then we went back to Ch. 24 (wave nature), we skipped 24.9. For 24.8 we
really didn't do any quantitative CAPA problems, but did spend some
lecture time on it (so I wouldn't ask any more than a basic qualitative
question about "soap bubbles" or "antireflective coating" etc. )

We skipped CH. 26 (We have used the formulas "E=mc^2" for a particle at
rest,  and "E=pc" for photons, from that chapter)

We did everything in Ch. 27 except 27-7. 

We'll end up covering 28.1-7.  You can tell from the last homework
where the emphasis on that quantum stuff will likely be (Mostly
Heisenberg uncertainty?). We haven't spent as much time on Ch. 28 as on
the others, so it won't be as prominant on the final.

The final will be more than half "new material". The rest will cover
everything from the whole semester.




 SO many questions!! 

>
>The relationship is d sin theta = m lambda
>        1. When lambda increases, d stays the same --- theta increases and
>interference spreads out, correct?      B/c sin theta = m lambda/d and if
>lambda increases and d doesn't' change then the angle would have to     get
>larger.

Yes. E.g. Red light on a given grating is spread out MORE than blue light.

>
>        2. If you have a big slit you have a smaller lambda, correct? Then

Lambda is up to you, it's whatever color you choose to shine on the slit!
You can shine ANY lambda on any slit..

>you
>won't see a diffraction pattern --      b/c the pattern will be too small?

But yes, if you choose to shine a small lambda on a big slit, the light
behaves like you'd expect geometrically, the wave nature is not
apparent. (The diffraction lines are so closely spaced they merge
together) You see a normal "shadow" of the slit...

>
>        3. If lambda is larger than the slit then you only see a broad
>pattern?

Yes.

d sin(theta) = m lambda, so if lambda is large, theta gets large. If
lambda is larger than d, then you can't even solve this equation for
any theta (because sin(theta) is never larger than 1!) so your pattern
has become so broad you don't even get to the first maximum...

>
>        4. When you get to electrons and the like - detection of interference
>only
>occurs in the slit distance is  less than or equal to  lambda, correct?
                    

It's not quite so formal as all this. You can see interference if you
look carefully enough even when lambda is much smaller than d, but it
just gets EASIER and more noticeable to see interference as lambda get
bigger. The "generic scale" where it's quite visible without any fancy
tricks is when lambda is about the same order of magnitude as the slit
size...


>Correspondingly in order to detect an electron (etc) the speeds         would
>have
>to be very slow in order to achieve a detectable lambda, correct?

? You can always detect an electron, independent of the speed. If you
want to see electrons INTERFERING with one another (i.e if you want to
observe the "wave nature" of electrons), e.g. giving you some
bizarre "minima" where no electrons go, then you want lambda to be
comparable to the slit spacing, which generally means pretty slow. (Not
THAT slow though, because electrons are very light)

>
>        5. Single slit grating:
>        * slit gets big - pattern looks like a shadow of a slit         (how
>        * big        

No definite answer to "how big", generically, when lambda is "much
less" than the slit opening.

>* Slit gets small - very wide pattern
>        * (how small is small?)

As usual, no definite answer, but small means "comparable to lambda".

>        * fixed slit, longer lambda to shorter lambda - pattern squeezes in
>        (why?)

First minimum (which in a real sense defines "how broad the pattern looks" is
Dsin(theta) = lambda,  or if you prefer, D (x_1st min /L) = lambda

So, if lambda is bigger, x is bigger, i.e. the pattern is broader. 
As lambda gets smaller, x gets smaller, the pattern "squeezes in".  

That's the mathematical answer. The physical answer is that the SMALLER
the slit, the more POINTLIKE the source, and pointlike sources send
waves out in ALL directions including very large angles.  (Huygens says
a point source makes a perfectly spherical wave)

Bottom line: small slits diffract light MORE strongly.

>        * fixed slit, shorter lambda to longer lambda - pattern spreads out
>        (why?)

Same as above...

>
>2. Are you going to cover telescopes on the final? In any event, the angular
>magnification for a telescope is M = theta prime/theta = - Fo/Fe.....The M =
>N/f only applies to magnifying glasses right?

Yes, I will, and yes, that's right.

>
>3. Lecture dated 4/12 - At the very beginning of class you drew the double
>slit diffraction grating.  You were talking about the CD and a prism and
>their differences.  You said they work the opposite way.  Can you explain
>that.  I understand the prism pretty well - it's the CD....  You also said
>if m = 0 then there is no separation of color....Can you explain.

Diffraction gratings of any kind (slits, CD's, whatever) spread out red
light MORE, to larger angles. It's from the formula 
d sin(theta) = m lambda, 
so if lambda is bigger (redder), theta is bigger (more deviation, 
more spreading)

Prisms tend to shift the direction of red a little, and the blue MORE.
That's an *accident* of nature, the fact the index of refraction in
many (not all) materials is slightly larger for shorter wavelengths.
(So blue sees a slightly *larger* index than red does, which tends to
refract the blue light MORE strongly)

If m=0, dsin(theta)=0*lambda =0. So ANY wavelength light will travel
through in a straight line (theta=0 is the solution to that equation!)
So white light doesn't get clor separated, ALL the wavelengths travel
exactly the same into the m=0 line...


>
>4. Double Slit Grating:
>I have in my in notes 1/2x = d sin theta = 1/2 lambda - I don't understand
>this - would this be the point of a dark spot?......Wouldn't x be (m *
>lambda * L)/d. And we can use this to when the angle is less than 10
>degrees, correct?  If the angle is larger than 10 degrees we go back to
>using sin, correct?

I'm not sure I know what that formula is either, I don't understand the
"1/2x" at the left. The formula d sin(theta)= 1/2 lambda would be the
condition for a minimum (dark spot) for a double slit.  (Perhaps you
were making notes for single slit? Even so, I'm not 100% sure what that
formula is referring to)

Around 10 degrees is indeed the cutoff for using the small angle
approximation, (sin(theta)=tan(theta)=theta = x/L, in which case 
x = m lambda *L/d for bright spots in a double or multiple slit example.)


>
>5. Can you define diffraction limit.

When light goes through any slit, it diffracts. The "diffraction limit"
for a telescope refers to the SMALLEST angle you can resolve, that
means the smallest angular separation two point objects can have and
you can still TELL there are two objects. If they get closer, to LESS
than your diffraction limit, then the diffractive "smearing" of the two
images make them overlap into one big fuzzy blob, and you can't tell
there are two things there.

The (rough) formula is angular separation must be greater than
lambda/D, where D is the diameter of the telescope.

>
>6. Page 24-19 of your notes - You don't talk about the relationship of n2 to
>n1 for say oil on water - the book on page 741 goes into detail about this -
>do we have to be able to do this?

I probably won't be asking any quantitative questions about "thin film"
interference, if anything just something qualitative relating to the
"bubble" demo or something. I don't think I assigned any CAPA questions
on thin films..

>
>7. Polarization: Say I have three polarizers - after the first polarizer I
>have Io/2, then the second polarizer is at 60 degrees so I have (Io/2) cos^2
>60, then the third polarizer is at 20 degrees so finally I have (Io/2 cos^2
>60) cos^2 20.....I guess the question is - as you progress through
>polarizers what you add to the next equation is the cos^2 angle of the new
>polarizer, correct?

Yes, cos^2(angle of the new polarizar WITH RESPECT TO THE PREVIOUS
ONE), the angle is always the angle between "direction of polarization
of incoming light" and "direction of new polarizer". (So, theta is NOT
the angle with respect to, say, some fixed direction...) I think you
understood this though. So in your case, if the first polarizer was
vertical, the second was 60 degrees from vertical, the third one must
have been 20 degrees FROM THAT LAST ONE (i.e. either 40 degrees, or 80
degrees, from vertical, I can't tell from what you told me). Make
sense?

>
>8. Page 27-11 of your notes.  The whole point is that momentum of photons is
>quantized, correct?  I guess also conservation of energy applies and the
>only thing that changes is the frequency  of the photon and consequently the
>lambda of the photon, correct?  This is non-classical b/c classical says the
>frequency wouldn't change...ie wave in wave out....Right.  This section
>isn't coming together for me.....Also the last statement on that page -
>conservation of momentum - you can predict angles - the angels generated by
>the new frequency?    So if you used P = mv instead of using P - E/c then
>you would not be able to predict the angles, correct?

Momentum is "quantized" in the sense that p = h/lambda is fixed (given
the color, i.e. given lambda).  When light scatters from an electron,
it gives energy to the electron (the electron recoils with KE, after
all), so the light must LOSE energy. But light cannot slow down, E =
hf, so if the photon loses energy, its frequency must go down. That's
conservation of energy. Yes, classical says frequency doesn't change,
color in = color out...

Conservation of momentum says p_init = p_final. p is a vector, it has x
and y components. Each is *separately* conserved. So if you measure the
energy and angle of, say, the recoiling electron, you know exactly its
x and y momenta components, and then the energy and ANGLE of the
recoiling photon would be completely determined by conservation of
energy, and p_x, and p_y. I wouldn't ask any quantitative questions
about this, but would expect you to understand qualitative aspects
like: the more energy the electron gets, the longer the wavelength of
the final photon, stuff like that.

(If you tried to use *non*relativistic kinematics for the photon, none
of the details of the prediction of angle above would work)


>
>In my notes before the photon crashes into the electron ( I have x-rays
>coming in and longer waves coming out) I have E(photon) and P(photon) coming
>in --- Is this supposed to represent E * P? Or just that they are both
>coming in with the photon?  I think they are coming in separately.

They are separate, different things. A photon has E=hf, and
p=h/lambda.  Two separate properties, just like any particle has
energy, and momenta. They are separate, but *related*, by E = |p c|.
(For ordinary nonrelativistic particles, the connection would be E =
p^2/2m)

>
>9. Relativistic:  light c = v
>E = hf for one photon
>P = E/c = h/lambda
>
>Non-relativistic (v< c)
>E = 1/2 mv^2
>lambda = h/P = h/mv
>
>But for relativistic lambda also = h/mv....
 
No, it doesn't! A photon has m=0, and v=c. You can't use lambda=h/mv for
photons.

>So the difference is that for
>relativistic c = v and for non-relativistic you need to determine v which
>should be less than c....Right? 

No, we really haven't covered what to do if v is VERY close to c. Just use
the formulas E=hf and P=E/c = h/lambda for photons and you'll be fine.

>How much less?  Is non- relativistic v = 0 m/s to but not 
>including 3e^8????

No sharp cutoff, but generally any speed less than 1% of c can be
safely considered "non-relativistic"

>
>10. On your notes page 27-18 - the predicted size of atoms is 1/2 angstrom?

Yes, I blew it TWICE in my notes, claiming the size of atoms was 5E-10,
the correct number is .5E-10

>
>Thanks

Phew! Hope this helps!
Cheers,
Steve P.

 More questions, this time on quantum stuff 

>
>Yesterday in class when we started talking about Bohr and r * p = Hbar *
>n - You said if you speed up, you double ________???

I'm not sure what I said. Here's the bottom line: if you go to higher n
in the Bohr model,  it's complicated!  Radius increases (r = n^2 * 0.5
Angstroms), energy increases, but in a funny way, it becomes LESS
negative. Velocity you can get from r*p = n hbar (that's the
quantization of angular momentum formula), so m v = n hbar / r, and
then you have to plug in for r = n^2*0.5, to get m v = hbar / n * 0.5
Angstroms (notice the n in the numerator cancels one n in the
denominator, but there's one left behind). So if n increases, velocity
actually DECREASES. (Far away orbits, the electron is moving more
slowly)


>
>Your notes Page 28-04 - Do you mean that you can't locate things if
>lambda >> objects size b/c long wavelengths mean bad resolution? If not,
>then what?

Yes, it's hard to look at small objects using long wavelengths.  Long
wavelengths have worse resolution is a fine way to put it. Generally, if
lambda is much bigger than the size of the object, the wave doesn't even
NOTICE the object! Visible light is useless for looking at things smaller
than a few thousand angstroms. To look at ATOMS, e.g, you need high energy
x-rays (or else, an electron microscope with lambda_de Broglie small) 


>
> What is the quantum limit - I mean I understand what it is -but is there
>a value (numerical) to it?

No value. It depends on the situation. Generally, atomic sized things
are "quantum". (But e.g. Prof Wieman over in the Jila tower has managed
to create a macroscopic (mm sized?) chunk of "quantum matter", a
Bose-Einstein condensate, in his lab!)  Generally, when the product of
sizes and momenta get close to the HUP limit, x*p around hbar, life
gets "quantum-y".


 Ooops! 
>When do you use E=kQ/r^2 as opposed to E=kQ/r? Exam 1,001 uses kQ/r for
>question 4. 

E=kQ/r^2 is *always* correct. That was a typo in my solution! You're
the first person to notice it!

The solution is still essentially correct, fortunately: it didn't rely
on any number crunching (if r is smaller, 1/r is bigger, and 1/r^2 is
bigger still, so the qualitative nature of the problem didn't care
whether the formula was 1/r or 1/r^2)

But anyway, E=kQ/r^2, absolutely.

(The formula kQ/r is used for the electrical *potential*, not the
electric field.)











  

    
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