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Chapter 11: Problem 20

A beam of unpolarized light is passed through two polarizers. If the polarization axis of the second polarizer is at an angle of \(45^{\circ}\) with respect to the axis of the first polarizer, then the intensity of light seen by someone located to the right of the second polarizer is (A) the intensity of the original light. (B) one-half the intensity of the original light. (C) one-quarter the intensity of the original light. (D) one-eighth the intensity of the original light. (E) zero.

Short Answer

Expert verified
Question: After an unpolarized light beam passes through two polarizers set at an angle of 45 degrees with respect to each other, the intensity of the transmitted light is: A) one-eighth the intensity of the original light B) one-half the intensity of the original light C) one-quarter the intensity of the original light D) the same as the intensity of the original light Answer: C) one-quarter the intensity of the original light.

Step by step solution

01

Identify given information

We are given that the angle between the two polarizers is \(45^{\circ}\) (or \(\frac{\pi}{4}\) radians).
02

Apply Malus' Law for the first polarizer

When light passes through the first polarizer, it gets polarized, and its intensity is reduced by half. Therefore, the intensity of the polarized light (I_polarized) after the first polarizer is: \(I_{polarized} = \frac{1}{2}I_0\) Where \(I_0\) is the intensity of the original (unpolarized) light.
03

Apply Malus' Law for the second polarizer

The intensity of the light (I_transmitted) transmitted through the second polarizer can be found using Malus' Law: \(I_{transmitted}=I_{polarized}\cos^2{\theta}\) Where \(\theta\) is the angle between the polarizers, which is given as \(45^{\circ}\).
04

Calculate the transmitted light's intensity

Plug in the value of \(\theta\) and the intensity after passing through the first polarizer: \(I_{transmitted} = \frac{1}{2}I_0 \cos^2 \left( \frac{\pi}{4} \right)\) We know that \(\cos^2(45^{\circ}) = \cos^2\left(\frac{\pi}{4}\right) = \frac{1}{2}\). Substitute this value: \(I_{transmitted} = \frac{1}{2}I_0 \cdot \frac{1}{2}\) \(I_{transmitted} = \frac{1}{4}I_0\)
05

Choose the correct answer

The transmitted light's intensity is one-quarter the intensity of the original light. Therefore, the correct answer is: (C) one-quarter the intensity of the original light.

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