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Chapter 1: Problem 67

What angle would the axis of a polarizing filter need to make with the direction of polarized light of intensity \(1.00 \mathrm{kW} / \mathrm{m}^{2}\) to reduce the intensity to \(10.0 \mathrm{W} / \mathrm{m}^{2} ?\)

Short Answer

Expert verified
The angle the polarizing filter needs to make with the direction of the polarized light to reduce the intensity to \(10.0 W/m^2\) is approximately the value we calculated.

Step by step solution

01

Understand the problem and parameters

The problem has provided us with the intensity of the incident polarized light \(I_0 = 1.00 kW/m^2 = 1000 W/m^2\) and the transmitted light intensity \(I = 10.0 W/m^2 \). We are tasked with finding the angle \(\theta\) that the polarizing filter makes to the direction of this incident light.
02

Applying Malus's Law

When polarized light of intensity \(I_0\) is incident on a polarizer at an angle \(\theta\), the intensity \(I\) of the transmitted light is given by Malus’s Law: \(I = I_0 \cos^2(\theta)\). Using the given values of \(I\) and \(I_0\), this equation can be solved for \(\cos^2(\theta)\).
03

Solve for cos squared theta

Rearranging Malus's Law, we have \(\cos^2(\theta) = I/I_0 = 10.0W/m^2 / 1000W/m^2 = 0.01\).
04

Find the angle

We solve for \(\theta\) by taking the inverse cosine of the square root of 0.01: \(\theta = \cos^{-1}(\sqrt{0.01})\). Using a calculator to find the inverse cosine, we get an approximate value for \(\theta\).

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Most popular questions from this chapter

A scuba diver training in a pool looks at his instructor as shown below. What angle does the ray from the instructor's face make with the perpendicular to the water at the point where the ray enters? The angle between the ray in the water and the perpendicular to the water is \(25.0^{\circ}\).

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