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conantee
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« on: January 06, 2010, 01:37:42 PM »

74. Unpolarized light is incident on two ideal polarizers in series. The polarizers are oriented so that no light emerges through the second polarizer. A third polarizer is now inserted between the first two and its orientation direction is continuously rotated through 180^o. The maximum fraction of the incident power transmitted through all three polarizers is
(A) zero
(B) \frac{1}{8}
(C) \frac{1}{2}
(D) \frac{1}{\sqrt{2}}
(E) 1
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« Reply #1 on: October 13, 2010, 08:57:06 AM »

The key here is property of polarizers that treats incoming light as vector. That is, the polarizer only allows the component of E-field parallel to the axis of polarizer to pass through. That's why when a third polarizer is inserted, the intensity of transmitted light can be changed.

Let me illustrate by symbols. Suppose the first polarizer aligns vertically, \updownarrow  , and the second polarizer aligns horizontally, \leftrightarrow. If there is no third polarizer. The transmitted light from the first polarized will be always blocked by the second polarizer since the transmitted field is perpendicular to the axis of the second polarizer. However, if the third polarizer aligns with inclination, the transmitted field from the 1st polarizer will not purely perpendicular to the axis of the 3rd one. Then, there will be transmitted light from the 3rd polarizer, and the transmitted light will not purely perpendicular to the axis of the 2nd one. As a result, there is some transmitted light from the 2nd polarizer.

What is the condition to maximize the intensity then? If you remember Malus law, that's great. If not, argue by symmetry. We can do the experiment backward by sending light to the 2nd polarizer first. So, the angle between the 3rd polarizer and the 1st polarizer should be equal to the angle between the 3rd and the 2nd - that is 45^o. Undergoing the projection from the 1st to 3rd to 2nd, the field strength decreases by factor \cos(45^o) \cos(45^o) = 1/2. We know that intensity is proportional to the field squared. Hence, through projection, the intensity of light will be reduced by at least factor of 4. Hence, (C), (D) and (E) are wrong, and we know that the intensity left is not zero. So, the answer is (B).

The extra factor of 2 comes from the fact that the light source is unpolarized, so the intensity of light is reduced by the first polarizer since the component perpendicular to the axis of polarizer is blocked. By randomness between perpendicular case and parallel case, the intensity left is reduced by half.

Observe that you can do this question without concrete knowledge of what is actually going on.

The correct answer is (B). 30 of 100 got this one right.
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