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## GR0877 Problem 16

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (D) When unpolarized light passes through the first polarizer it loses one-half of its intensity. Why a half? Here is an explanation. A beam of unpolarized light is nothing but a uniform mixture of linear polarizations at all possible angles. When a polarized beam of light of intensity $I_0$ is incident on a perfect polarizer, the intensity of light $I$ that passes through is $I = I_0 \cos^2 \theta$ (Malus’s law). Averaging this over all possible angles $\theta$ and using the fact that the average of $\cos^2 \theta$ is 1/2 (to prove, average the identity $\sin^2 \theta + \cos^2 \theta =1$ over one cycle), one has: $I_1 = I_0/2$, where $I_1$ is the intensity of light after the first polarizer. After the first polarizer the light is linearly polarized and the fraction of light that passes through the second polarizer is given by (again, using Malus’s law): $I_2 = I_1 \cos^2 45^{\circ} = I_0/4$ or 25%.

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Categories: 0877 Tags: , , , ,
1. 29.08.2012 at 01:40

The 5th sentence, should it read, “When a perfect polarizer is placed in an unpolarized beam….”? rather than a polarized beam?

• 29.08.2012 at 22:55

NO. According to Malus’ law, when plane POLARIZED light is incident on the polarizer, the intensity of the light transmitted is directly proportional to the square of the cosine of corresponding angle. Now, we have UNpolarized light which is incident on the (first) polarizer and we want to know what fraction of this light is transmitted through the polarizer. We deduce this using Malus’ law for POLARIZED light, noting that UNpolarzied light is a uniform mixture of linear polarizations at all possible angles (we take average). After the first polarizer light is POLARIZED and we again use Mallus’ law to get the answer.

Hope this helps.

• 29.08.2012 at 23:39

Ok, I gotcha now. Thanks.

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