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Chapter 29: Problem 42

Unpolarized light of intensity \(S_{0}\) passes first through a polarizer with its axis vertical and then through one with its axis at \(35^{\circ}\) to the vertical. Find the intensity after the second polarizer.

Short Answer

Expert verified
The intensity of the light after passing through the two polarizers is approximately \( 0.4095 \cdot S_{0} \).

Step by step solution

01

Light Passing Through First Polarizer

When unpolarized light passed through a first polarizer, it becomes polarized and the intensity of the light is halved. Let's denote the intensity after passing the first polarizer as \( S_{1} \). Using the principle of polarization, we can write the intensity as \( S_{1} = S_{0}/2 \).
02

Light Passing Through Second Polarizer

When the pre-polarized light \( S_{1} \) passes through a second polarizer that is oriented at an angle \( \theta = 35^{\circ} \) from the first polarizer. We can make use of Malus' Law to find the transmitted light intensity \( S_{2} \) as follows: \( S_{2} = S_{1} \cdot \cos^{2}{\theta} = S_{0}/2 \cdot \cos^{2}{35^{\circ}} \).
03

Simplify the Result

To simplify, remember that \( \cos^{2}{35^{\circ}} \) is approximately \( 0.819 \). Plug this and simplify to get the result: \( S_{2} \approx 0.819 \cdot S_{0}/2.

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