GR9277 #67
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Alternate Solutions |
oliTUTilo 2012-11-04 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 (A-B), we must have one plane-polarized 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 plane-polarized can be described as a sum of partly plane-polarized 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 plane-polarized and partly unpolarized, as in choice (C). | |
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Comments |
oliTUTilo 2012-11-04 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 (A-B), we must have one plane-polarized 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 plane-polarized can be described as a sum of partly plane-polarized 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 plane-polarized and partly unpolarized, as in choice (C). | | swepi0 2010-11-11 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 2012-10-02 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)?
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| | narfle 2010-10-02 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 2010-11-11 21:34:29 |
Remember that A>B whereas the trig identity
leaves us with a constant that satisfies
< for some values of .
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| | jmason86 2009-09-07 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 2009-09-28 13:42:50 |
an elliptically polarized wave.
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kroner 2009-09-28 17:04:54 |
That is assuming the light is in a coherent plane wave. It doesn't have to be.
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Munin 2010-04-09 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.
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| | neoslovakia 2007-10-22 06:45:41 | It doesnt matter for this problem, but you guys have the trig identity wrong. It should say
| | Richard 2007-10-18 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 2010-10-22 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.
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checkyoself 2011-10-04 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.
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| | radicaltyro 2006-10-31 11:01:35 | What about circular polarization?
Shoshe 2006-11-03 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 plane-polarised part of the light has the oscillating intensity .
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| | jax 2005-12-06 08:22:42 | In the exam I am looking at, it says for the plane polarized part not . Is that a mistake?
yosun 2005-12-06 20:07:14 |
jax: recall the trig identity .
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mhas035 2007-04-05 14:41:43 |
Or
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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 (A-B), we must have one plane-polarized 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 plane-polarized can be described as a sum of partly plane-polarized 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 plane-polarized and partly unpolarized, as in choice (C).
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