/*! 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 97 Two polarizing sheets, one direc... [FREE SOLUTION] | 91Ó°ÊÓ

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Two polarizing sheets, one directly above the other, transmit \(p \%\) of the initially unpolarized light that is perpendicularly incident on the top sheet. What is the angle between the polarizing directions of the two sheets?

Short Answer

Expert verified
The angle is \( \theta = \cos^{-1}\left(\sqrt{\frac{p}{50}}\right) \).

Step by step solution

01

Understanding the Problem

We have two polarizing sheets, and they transmit a certain percentage of unpolarized light. When light passes through a polarizer, the intensity follows Malus's Law. Our task is to find the angle between the polarizing directions of these sheets given the transmission percentage.
02

Applying the Law of Malus

When unpolarized light passes through the first polarizer, its intensity is reduced by half. The intensity after the first sheet is \( \frac{I_0}{2} \). Let \( \theta \) be the angle between the polarization directions of the two sheets. According to Malus's Law, the transmitted intensity \( I \) after the second polarizer is \( I = \frac{I_0}{2} \cos^2(\theta) \).
03

Express the Given Percentage

We need to express that the transmitted intensity \( I \) is \( p\% \) of the original intensity \( I_0 \), i.e., \( I = \frac{p}{100} I_0 \). Therefore, we equate \( \frac{I_0}{2} \cos^2(\theta) = \frac{p}{100} I_0 \).
04

Solve for \( \cos^2(\theta) \)

Simplify the equation by canceling \( I_0 \) from both sides, leading to \( \frac{1}{2} \cos^2(\theta) = \frac{p}{100} \). Solve for \( \cos^2(\theta) \): \( \cos^2(\theta) = \frac{2p}{100} = \frac{p}{50} \).
05

Determine Angle \( \theta \)

Take the square root of both sides to find \( \cos(\theta) \), giving us \( \cos(\theta) = \sqrt{\frac{p}{50}} \). Find \( \theta \) using the inverse cosine function: \( \theta = \cos^{-1}\left(\sqrt{\frac{p}{50}}\right) \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Malus's Law
Malus's Law is an important principle in the study of polarized light, named after the French physicist Étienne-Louis Malus. This law describes how the intensity of polarized light changes as it passes through another polarizing material. According to Malus's Law, when light that is initially polarized by one polarizer passes through a second polarizer, the intensity of the light is given by the equation: \[ I = I_0 \cos^2(\theta) \]where:
  • \( I \) is the intensity of the light after passing through both polarizers.
  • \( I_0 \) is the initial intensity of the light before the second polarizer.
  • \( \theta \) is the angle between the transmission axis of the first and second polarizer.
This relationship shows that the transmitted light's intensity depends on the cosine of the angle squared, meaning as the angle changes, the amount of light getting through drastically varies. When \( \theta \) is 0 degrees, all polarized light passes through; when \( \theta \) is 90 degrees, no light passes through. Understanding Malus's Law is fundamental in applications involving polarized light, such as in sunglasses and photographic filters.
Polarizing Sheets
Polarizing sheets, also known as polarizers, are materials that allow light waves of only a certain polarization to pass through. They function by blocking certain orientations of light waves while letting others continue. When unpolarized light, which consists of waves vibrating in all possible directions perpendicular to the direction of propagation, encounters a polarizing sheet, the sheet selectively filters these waves:
  • The first polarizer reduces the intensity of light by allowing only waves aligned with its axis to pass.
  • Subsequent polarizers can further reduce intensity depending on their orientation relative to the first.
Polarizing sheets are widely used in many technologies including LCD screens, camera lenses, and in reducing glare in sunglasses. They are integral in experiments and devices designed to examine the properties and behavior of light.
Intensity of Light
The intensity of light refers to the power per unit area carried by a wave. When dealing with polarizers, intensity becomes especially important since it dictates how much light ultimately passes through the system of sheets.For unpolarized light, when it encounters a polarizer, the intensity is initially halved because the sheet only allows through the component of light aligned with its axis. This results in:\[ I_1 = \frac{I_0}{2} \]for the light passing through the first polarizer, where \( I_0 \) is the original intensity of unpolarized light.Upon passing through a second polarizer, Malus's Law comes into play, and the resulting intensity is:\[ I = \frac{I_0}{2} \cos^2(\theta) \]Understanding how intensity alters through a series of polarizers helps in solving problems involving light transmission and in designing systems where controlling light brightness is crucial. This understanding is key in optical equipment like microscopes and in analytical techniques.

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Most popular questions from this chapter

Light in vacuum is incident on the surface of a glass slab. In the vacuum the beam makes an angle of \(32.0^{\circ}\) with the normal to the surface, while in the glass it makes an angle of \(21.0^{\circ}\) with the normal. What is the index of refraction of the glass?

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