GR9277 #67



Alternate Solutions 
oliTUTilo 20121104 13:18:16  I think that completely unpolarized light and circularly polarized light would behave the same on our filter system. That is, they would both transmit a constant intensity over filter angles.
Completely unpolarized light has components of polarization in all directions.
Circularly polarized light has two components perpendicular to each other and a quarter out of phase, creating a net rotating (or circular) polarization. But at low speeds of rotations circularly polarized light transmits polarizations of all orientations with respect to the filter.
Since there is a term of intensity that is proportional to (namely ), and one term that is a constant (AB), we must have one planepolarized component and one that is either unpolarized or circularly polarized, given that we can only choose up to two types of components. It turns out that all light that is not completely planepolarized can be described as a sum of partly planepolarized and partly unpolarized light. Since we can't tell from the intensity formula if the constant intensity component comes from circularly polarized light or unpolarized light, we can at least conclude that the light is partly planepolarized and partly unpolarized, as in choice (C).  

Comments 
oliTUTilo 20121104 13:18:16  I think that completely unpolarized light and circularly polarized light would behave the same on our filter system. That is, they would both transmit a constant intensity over filter angles.
Completely unpolarized light has components of polarization in all directions.
Circularly polarized light has two components perpendicular to each other and a quarter out of phase, creating a net rotating (or circular) polarization. But at low speeds of rotations circularly polarized light transmits polarizations of all orientations with respect to the filter.
Since there is a term of intensity that is proportional to (namely ), and one term that is a constant (AB), we must have one planepolarized component and one that is either unpolarized or circularly polarized, given that we can only choose up to two types of components. It turns out that all light that is not completely planepolarized can be described as a sum of partly planepolarized and partly unpolarized light. Since we can't tell from the intensity formula if the constant intensity component comes from circularly polarized light or unpolarized light, we can at least conclude that the light is partly planepolarized and partly unpolarized, as in choice (C).   swepi0 20101111 21:56:37  Correction:
I meant the identity
so that we will need some unpolarized light in order that A>B, since A=B if the light is only plane polarized.
hjq1990 20121002 22:55:14 
hello guys what if A=3*B? Then A+Bcos^2\theta=3B+b(2cos^\theta+1)=2B(1+cos^2\theta)?

  narfle 20101002 08:31:58  Why couldn't it just be B  just plane polarized. As we've seen, the trig identity applied to polarization creates a "1" as a constant  and there's your A=Bcos (2theta) right there without non polarized addition
swepi0 20101111 21:34:29 
Remember that A>B whereas the trig identity
leaves us with a constant that satisfies
< for some values of .

  jmason86 20090907 13:00:45  How can you have a wave that is partly polarized and partly unpolarized? That makes no sense to me when I think about wave shapes. Maybe it fits the equation, the unpolarized part will always get its intensity reduced to 1/2 and the plane polarized will be affected differently depending on ... so the equation works.
But what the crap does this wave look like?
kroner 20090928 13:42:50 
an elliptically polarized wave.

kroner 20090928 17:04:54 
That is assuming the light is in a coherent plane wave. It doesn't have to be.

Munin 20100409 19:57:04 
Just think of it as the superposition of two waves one that is polarized and one that is partly polarized add these together (think a circle and a ellipse) and you will get the combined wave.

  neoslovakia 20071022 06:45:41  It doesnt matter for this problem, but you guys have the trig identity wrong. It should say
  Richard 20071018 21:14:01  I think there is some sentiment for more pedentry here:
According to Malus' Law, circularly polarized light has intensity and unpolarized light is .
Using the trig identity , the intensity of circular polarized light is expressible as
So if we add in unpolarized light it is,
flyboy621 20101022 15:39:21 
So if A and B were equal, the incident light would be circularly polarized. Since we are given that A > B, there must be an additional unpolarized component to the incident light. Hence (C) is correct.

checkyoself 20111004 00:00:42 
I agree but surely in that case the answer is D? We start off with circular polarisation, identify that there's an unpolarised part due to the constant and we have answer D. I know the answer is C from ets so can someone shed light on this please.

  radicaltyro 20061031 11:01:35  What about circular polarization?
Shoshe 20061103 15:18:36 
If the incident light was completely circularly polarised, the intensity of the light passing through the rotating polaroid would be constant. The circularly polarised part of the light has transmitted intensity , and the planepolarised part of the light has the oscillating intensity .

  jax 20051206 08:22:42  In the exam I am looking at, it says for the plane polarized part not . Is that a mistake?
yosun 20051206 20:07:14 
jax: recall the trig identity .

mhas035 20070405 14:41:43 
Or

 




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