/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 51 Two sheets of polarizing materia... [FREE SOLUTION] | 91影视

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Chapter 22: Problem 51

Two sheets of polarizing material are placed with their transmission axes at right angles to one another. A third polarizing sheet is placed between them with its transmission axis at \(45^{\circ}\) to the axes of the other two. (a) If unpolarized light of intensity \(I_{0}\) is incident on the system, what is the intensity of the transmitted light?(b) What is the intensity of the transmitted light when

Short Answer

Expert verified
Answer: The intensity of the transmitted light with three sheets is \(\frac{I_{0}}{4}\), and the intensity of the transmitted light with two sheets is \(0\).

Step by step solution

01

Identify the angles between the sheets

As the third sheet is placed at \(45^{\circ}\) to both the first and second sheets, we know that: - Angle between Sheet 1 and Sheet 3: \(\theta_{1} = 45^{\circ}\) - Angle between Sheet 3 and Sheet 2: \(\theta_{2} = 45^{\circ}\)
02

Apply Malus Law for Sheet 1 and Sheet 3

We'll first find the intensity after Sheet 3 using the input light intensity \(I_{0}\): $$I_3 = I_0 \cos^2(\theta_1) = I_0 (\cos 45^{\circ})^2 = I_0 (\frac{\sqrt{2}}{2})^2 = \frac{I_0}{2}$$
03

Apply Malus Law for Sheet 3 and Sheet 2

Now, we'll find the transmitted intensity after Sheet 2 using the intensity after Sheet 3: $$I_t = I_3 \cos^2(\theta_2) = \frac{I_0}{2} (\cos 45^{\circ})^2 = \frac{I_0}{2} (\frac{\sqrt{2}}{2})^2 = \frac{I_0}{4}$$ So the intensity of the transmitted light when all three sheets are present is \(\frac{I_{0}}{4}\). ##Step 2: Calculating the intensity for Scenario B with only two polarizing sheets ##
04

Identify the angle between the remaining sheets

Since the sheets are at right angles to one another, the angle between Sheet 1 and Sheet 2 is: $$\theta_{3} = 90^{\circ}$$
05

Apply Malus Law for Sheet 1 and Sheet 2

Now, we'll find the transmitted intensity when the third sheet is removed: $$I_t' = I_0 \cos^2(\theta_3) = I_0 (\cos{90^{\circ}})^2 = I_0 (0)^2 = 0$$ So the intensity of the transmitted light when the third sheet is removed is \(0\). In conclusion, the intensity of the transmitted light with three sheets is \(\frac{I_{0}}{4}\), and the intensity of the transmitted light with two sheets is \(0\).

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