Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 22: Problem 53

Just after sunrise, you look north at the sky just above the horizon. Is the light you see polarized? If so, in what direction?

Short Answer

Expert verified
Answer: Yes, the light from the sky just above the horizon just after sunrise is polarized. It is polarized in the horizontal direction.

Step by step solution

01

Understanding the scattering of sunlight and polarization

Sunlight consists of unpolarized light. When it enters the Earth's atmosphere, it gets scattered by the air molecules, dust and aerosols particles. This process of scattering is highly dependent on the angle between the incident sunlight and the direction in which the scattered light is observed. The phenomenon responsible for the polarization of scattered light is known as Rayleigh scattering.
02

Principle of Rayleigh scattering and Brewster's angle

Rayleigh scattering asserts that the degree of polarization of scattered light depends on the angle between the incident light (sunlight) and the observer (us). At a specific angle, called the Brewster's angle, the degree of polarization is maximum, and the scattered light becomes completely polarized. Brewster's angle is given by the formula \(\theta_B = \cot^{-1} \left( \frac{n_2}{n_1} \right)\), where \(n_1\) and \(n_2\) are the refractive indices of the two media involved.
03

Determining the position of the sun and observer

Just after sunrise, the sun is approximately at the horizon in the east direction. When we look north, we are looking perpendicular to the direction of sunlight. So, the angle between the incident sunlight and the observer's direction is 90 degrees.
04

Evaluating the polarization in the given scenario

Since the angle between the incident sunlight and the observer's direction is 90 degrees, and considering that a 90-degree angle is a typical Brewster's angle in air, we can conclude that the light from the sky just above the horizon is polarized.
05

Finding the direction of polarization

The direction of polarization is perpendicular to the plane containing the incident light and the scattered light. In this case, since we are looking north and the sunlight is coming from the east, the plane containing the incident light and the scattered light will be a vertical plane. Therefore, the direction of polarization will be horizontal. In conclusion, the light from the sky just above the horizon, just after sunrise, is polarized in the horizontal direction.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 magnetic dipole antenna is used to detect an electromagnetic wave. The antenna is a coil of 50 turns with radius \(5.0 \mathrm{cm} .\) The EM wave has frequency \(870 \mathrm{kHz}\) electric field amplitude $0.50 \mathrm{V} / \mathrm{m},\( and magnetic field amplitude \)1.7 \times 10^{-9} \mathrm{T} .$ (a) For best results, should the axis of the coil be aligned with the electric field of the wave, or with the magnetic field, or with the direction of propagation of the wave? (b) Assuming it is aligned correctly, what is the amplitude of the induced emf in the coil? (Since the wavelength of this wave is much larger than \(5.0 \mathrm{cm},\) it can be assumed that at any instant the fields are uniform within the coil.) (c) What is the amplitude of the emf induced in an electric dipole antenna of length \(5.0 \mathrm{cm}\) aligned with the electric field of the wave?
What is the speed of light in a diamond that has an index of refraction of \(2.4168 ?\)
Prove that, in an EM wave traveling in vacuum, the electric and magnetic energy densities are equal; that is, prove that,$$\frac{1}{2} \epsilon_{0} E^{2}=\frac{1}{2 \mu_{0}} B^{2}$$ at any point and at any instant of time.,
The solar panels on the roof of a house measure \(4.0 \mathrm{m}\) by $6.0 \mathrm{m} .\( Assume they convert \)12 \%$ of the incident EM wave's energy to electric energy. (a) What average power do the panels supply when the incident intensity is \(1.0 \mathrm{kW} / \mathrm{m}^{2}\) and the panels are perpendicular to the incident light? (b) What average power do the panels supply when the incident intensity is \(0.80 \mathrm{kW} / \mathrm{m}^{2}\) and the light is incident at an angle of \(60.0^{\circ}\) from the normal? (W) tutorial: solar collector) (c) Take the average daytime power requirement of a house to be about 2 kW. How do your answers to (a) and (b) compare? What are the implications for the use of solar panels?
Calculate the frequency of an EM wave with a wavelength the size of (a) the thickness of a piece of paper \((60 \mu \mathrm{m}),(\mathrm{b})\) a \(91-\mathrm{m}\) -long soccer field, (c) the diameter of Earth, (d) the distance from Earth to the Sun.
See all solutions

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Electric Charge Field and Potential

Read Explanation

Scientific Method Physics

Read Explanation

Rotational Dynamics

Read Explanation

Fields in Physics

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.

Sign-up for free