/*! 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 41 The average intensity of light e... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 24: Problem 41

The average intensity of light emerging from a polarizing sheet is \(0.764 \mathrm{W} / \mathrm{m}^{2},\) and the average intensity of the horizontally polarized light incident on the sheet is \(0.883 \mathrm{W} / \mathrm{m}^{2} .\) Determine the angle that the transmission axis of the polarizing sheet makes with the horizontal.

Short Answer

Expert verified
The angle is approximately \(21.57^{\circ}\).

Step by step solution

01

Understand the Problem

We need to find the angle the transmission axis of the polarizing sheet makes with the horizontal when given initial and emerging light intensities.
02

Utilize Malus's Law

Malus's Law gives the intensity of light after passing through a polarizing filter as: \[ I = I_0 \cos^2(\theta) \] where \( I_0 = 0.883 \, \mathrm{W/m^2} \) is the initial intensity, \( I = 0.764 \, \mathrm{W/m^2} \) is the emerging intensity, and \( \theta \) is the angle we want to find.
03

Rearrange Malus's Law

We rearrange the formula to find the angle: \[ \cos^2(\theta) = \frac{I}{I_0} \] Substituting the given values: \[ \cos^2(\theta) = \frac{0.764}{0.883} \] Calculate this fraction.
04

Calculate the Fraction

Divide the emerging intensity by the initial intensity: \[ \cos^2(\theta) = \frac{0.764}{0.883} \approx 0.865 \]
05

Solve for \(\theta\)

Take the square root of both sides to find \( \cos(\theta) \): \[ \cos(\theta) \approx \sqrt{0.865} \approx 0.930 \] Now, find \( \theta \) using the arccos function: \[ \theta = \cos^{-1}(0.930) \]
06

Compute the Angle

Use a calculator to find \( \theta \): \[ \theta \approx \cos^{-1}(0.930) \approx 21.57^{\circ} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Polarizing Sheet
A polarizing sheet, also known as a polarizer, is a crucial optical device used in various applications to filter light based on its polarization. Light waves can oscillate in different directions, and when light encounters a polarizing sheet, it only allows light with a specific polarization direction to pass through. This direction is aligned with the transmission axis of the polarizing sheet.

The transmission axis is essentially the line along which light waves are allowed to pass. When unpolarized light, containing waves oscillating in multiple directions, hits a polarizing sheet, it emerges as polarized light, with waves oscillating parallel to the transmission axis.

Polarizing sheets are widely used in sunglasses, cameras, and scientific instruments to manage glare and improve the visibility of certain features by filtering out undesired light directions.
Light Intensity
Light intensity refers to the amount of energy the light wave carries per unit area, often expressed in watts per square meter ( W/m^2 ). It's a key factor when evaluating how much energy light transmits when it passes through materials like polarizing sheets.

In the context of the given problem, two light intensities are of particular interest:
  • The initial intensity ( I_0 ), which represents the light's power before it enters the polarizing sheet.
  • The emerging intensity ( I ), which measures the light's power after it has passed through the polarizer.
The change in light intensity is dictated by Malus's Law, which mathematically relates these intensities to the angle of the transmission axis with respect to the incident light's polarization.
Angle of Polarization
The angle of polarization is the angle between the transmission axis of a polarizing sheet and the direction of the electric field of the incident light. In simpler terms, it's the angle at which the light is aligned with the transmission axis.

Determining this angle is crucial because it affects how much light is transmitted through the polarizer. Malus's Law mathematically expresses this relationship as:\[ I = I_0 \cos^2(\theta) \]where \(\theta\) is the angle of polarization. In problems like the one described, solving for this angle helps us understand the orientation of the polarizing sheet relative to the incident light.

As calculated using inverse cosine functions, knowledge of this angle helps in practical applications where controlling light intensity and directionality is important, such as in photography and optical engineering.
Transmission Axis
The transmission axis is a fundamental aspect of polarization that defines how light is transmitted through a polarizing sheet. It determines the specific direction along which light is allowed to pass, effectively filtering out all light vibrations not aligned with this axis.

When horizontally polarized light hits a polarizer, only the component of the light parallel to the transmission axis gets through. This makes the transmission axis paramount in applications needing precise control over light intensity. In the exercise, finding the angle between the transmission axis and the horizontal provides insights into the degree of alignment between the polarizer and incoming light.

This concept is vital in many technologies, from LCD screens to photography filters, demonstrating the crucial role of the transmission axis in enhancing clarity and reducing unwanted reflections.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Neil A. Armstrong was the first person to walk on the moon. The distance between the earth and the moon is \(3.85 \times 10^{8} \mathrm{m}\). Find the time it took for his voice to reach the earth via radio waves. (b) Someday a person will walk on Mars, which is \(5.6 \times 10^{10} \mathrm{m}\) from the earth at the point of closest approach. Determine the minimum time that will be required for a message from Mars to reach the earth via radio waves.

Light that is polarized along the vertical direction is incident on a sheet of polarizing material. Only \(94 \%\) of the intensity of the light passes through the sheet and strikes a second sheet of polarizing material. No light passes through the second sheet. What angle does the transmission axis of the second sheet make with the vertical?

A distant galaxy emits light that has a wavelength of \(434.1 \mathrm{nm}\). On earth, the wavelength of this light is measured to be \(438.6 \mathrm{nm}\). (a) Decide whether this galaxy is approaching or receding from the earth. Give your reasoning. (b) Find the speed of the galaxy relative to the earth.

Unpolarized light whose intensity is \(1.10 \mathrm{W} / \mathrm{m}^{2}\) is incident on the polarizer in Figure \(24.20 .\) (a) What is the intensity of the light leaving the polarizer? (b) If the analyzer is set at an angle of \(\theta=75^{\circ}\) with respect to the polarizer, what is the intensity of the light that reaches the photocell?

A future space station in orbit about the earth is being powered by an electromagnetic beam from the earth. The beam has a cross-sectional area of \(135 \mathrm{m}^{2}\) and transmits an average power of \(1.20 \times 10^{4} \mathrm{W} .\) What are the rms values of the (a) electric and (b) magnetic fields?
See all solutions

Recommended explanations on Physics Textbooks

Electricity

Read Explanation

Medical Physics

Read Explanation

Radiation

Read Explanation

Engineering Physics

Read Explanation

Scientific Method Physics

Read Explanation

Work Energy and Power

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.