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GR9277 #67
Problem
 GREPhysics.NET Official Solution Alternate Solutions
\prob{67}

A steady beam of light is normally incident on a piece of polaroid. As the polaroid is rotated around the beam axis, the transmitted intensity varies as $A+Bcos(2\theta)$, where $\theta$ is the angle of rotation, and A and B are constants with $A\gt B\gt 0$. Which of the following may be correctly concluded about the incident light?

1. The light is completely unpolarized
2. The light is completely plane polarized
3. The light is partly plane polarized and partly unpolarized.
4. The light is partly circularly polarized and partly unpolarized.
5. The light is completely circularly polarized.

Optics$\Rightarrow$}Polarized light

A plane-polarized wave has intensity $I\propto \cos^2\theta$, where $\theta$ is the angle from the wave to the polarization axis. (This is also known as Malus' Law.)

An unpolarized wave has intensity $I = const$.

Since ETS is generous enough to supply the intensity, one can easily deduce choice (C).

Alternate Solutions
 oliTUTilo2012-11-04 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 ${(cos\theta)}^2$ (namely $B(1+cos(2\theta))$), 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).Reply to this comment
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 ${(cos\theta)}^2$ (namely $B(1+cos(2\theta))$), 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
$\cos^2\theta = (1/2)(1 + \cos 2\theta)$
so that we will need some unpolarized light in order that A>B, since A=B if the light is only plane polarized.
 hjq19902012-10-02 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)?
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
 swepi02010-11-11 21:34:29 Remember that A>B whereas the trig identity $\cos 2\theta = (1/2)(1 +\cos^2\theta)$ leaves us with a constant that satisfies $1/2$ < $\cos^2\theta$ for some values of $\theta$.
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 $I_0$ and the plane polarized will be affected differently depending on $\theta$... so the equation works.
But what the crap does this wave look like?
 kroner2009-09-28 13:42:50 an elliptically polarized wave.
 kroner2009-09-28 17:04:54 That is assuming the light is in a coherent plane wave. It doesn't have to be.
 Munin2010-04-09 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.
neoslovakia
2007-10-22 06:45:41
It doesnt matter for this problem, but you guys have the trig identity wrong. It should say
$cos^2\theta = \frac{1+cos(2\theta)}{2}$
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 $I_{circ}=A\cos^{2}(\theta)$ and unpolarized light is $I_{unpol}=B$.
Using the trig identity $\cos^{2}(\theta)=1+\cos(2\theta)$ , the intensity of circular polarized light is expressible as
$I_{circ}=A+A\cos(\theta)$
So if we add in unpolarized light it is,
$A+B+A\cos(\theta)=C+A\cos(\theta)$
 flyboy6212010-10-22 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.
 checkyoself2011-10-04 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.
2006-10-31 11:01:35
 Shoshe2006-11-03 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 $A$, and the plane-polarised part of the light has the oscillating intensity $B \cos 2 \theta$.
jax
2005-12-06 08:22:42
In the exam I am looking at, it says $cos 2 \theta$ for the plane polarized part not $cos^2 \theta$. Is that a mistake?
 yosun2005-12-06 20:07:14 jax: recall the trig identity $\cos(a + b) = \cos a \cos b - \sin a \sin b$.
 mhas0352007-04-05 14:41:43 Or $\cos^2\theta=1+\cos(2\theta)$

Correction: I meant the identity $\cos^2\theta = (1/2)(1 + \cos 2\theta)$ so that we will need some unpolarized light in order that A>B, since A=B if the light is only plane polarized.
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