Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 32: Problem 44

A light ray of wavelength 700 . \(\mathrm{nm}\) traveling in air \(\left(n_{1}=1.00\right)\) is incident on a boundary with a liquid \(\left(n_{2}=1.63\right) .\) a) What is the frequency of the refracted ray? b) What is the speed of the refracted ray? c) What is the wavelength of the refracted ray?

Short Answer

Expert verified
Question: A light ray with an initial wavelength of 700 nm travels from air (refractive index 1.00) into a liquid (refractive index 1.63). Calculate the frequency, speed, and wavelength of the refracted ray in the liquid. Answer: The refracted ray has a frequency of \(4.286\times10^{14}\, \mathrm{Hz}\), a speed of \(1.840\times10^8\, \mathrm{m/s}\) in the liquid, and a wavelength of \(430\, \mathrm{nm}\) in the liquid.

Step by step solution

01

Calculate the initial speed and frequency of the light ray in air

Recalling that the refractive index is given by \(n = \frac{c}{v}\), where \(v\) is the speed of light in the medium. We can solve for the speed of light in air as: $$ v_{1} = \frac{c}{n_{1}} $$ Then, we can calculate the frequency of the light ray using the relationship \(v = f \lambda\), where \(f\) is the frequency, and \(\lambda\) is the wavelength: $$ f = \frac{v_{1}}{\lambda_{1}} $$
02

Calculate the speed of the refracted ray in the liquid

Using the refractive index \(n_{2}\), we find the speed of the refracted light ray in the liquid as: $$ v_{2} = \frac{c}{n_{2}} $$
03

Calculate the wavelength of the refracted ray in the liquid

Since the frequency is constant when light travels from one medium to another, we have: $$ f_{1} = f_{2} $$ And by using the relationship \(v = f \lambda\), we can calculate the wavelength of the refracted ray in the liquid as: $$ \lambda_{2} = \frac{v_{2}}{f_{2}} = \frac{v_{2}}{f_{1}} $$ Now let's plug in the given values and solve for the required quantities:
04

Plug in the given values and solve

From steps 1, 2, and 3, we have: $$ v_{1} = \frac{3.00\times10^8\, \mathrm{m/s}}{1.00} = 3.00\times10^8\, \mathrm{m/s} $$ $$ f = \frac{3.000\times10^8\, \mathrm{m/s}}{700\times10^{-9}\, \mathrm{m}} = 4.286\times10^{14}\, \mathrm{Hz} $$ $$ v_{2} = \frac{3.00\times10^8\, \mathrm{m/s}}{1.63} = 1.840\times10^8\, \mathrm{m/s} $$ $$ \lambda_{2} = \frac{1.840\times10^8\, \mathrm{m/s}}{4.286\times10^{14}\, \mathrm{Hz}} = 430\times10^{-9}\, \mathrm{m} $$ So, the refracted ray has: a) a frequency of \(4.286\times10^{14}\, \mathrm{Hz}\), b) a speed of \(1.840\times10^8\, \mathrm{m/s}\) in the liquid, c) a wavelength of \(430\, \mathrm{nm}\) in the liquid.

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

An object is located at a distance of \(100 . \mathrm{cm}\) from a concave mirror of focal length \(20.0 \mathrm{~cm}\). Another concave mirror of focal length \(5.00 \mathrm{~cm}\) is located \(20.0 \mathrm{~cm}\) in front of the first concave mirror. The reflecting sides of the two mirrors face each other. What is the location of the final image formed by the two mirrors and the total magnification by the combination?

The shape of an elliptical mirror is described by the curve \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,\) with semi major axis \(a\) and semi minor axis \(b\). The foci of this ellipse are at points \((c, 0)\) and \((-c, 0)\) with \(c=\left(a^{2}-b^{2}\right)^{1 / 2}\). Show that any light ray in the \(x y\) -plane, which passes through one focus, is reflected through the other. "Whispering galleries" make use of this phenomenon with sound waves.

You are submerged in a swimming pool. What is the maximum angle at which you can see light coming from above the pool surface? That is, what is the angle for total internal reflection from water into air?

You have a spherical mirror with a radius of curvature of \(+20.0 \mathrm{~cm}\) (so it is concave facing you). You are looking at an object whose size you want to double in the image, so you can see it better. Where should you put the object? Where will the image be, and will it be real or virtual?

32.1 Legend says that Archimedes set the Roman fleet on fire as it was invading Syracuse. Archimedes created a huge ______ mirror, and he focused the Sun's rays on the Roman vessels. a) plane c) parabolic focusing b) parabolic diverging
See all solutions

Recommended explanations on Physics Textbooks

Nuclear Physics

Read Explanation

Astrophysics

Read Explanation

Classical Mechanics

Read Explanation

Magnetism

Read Explanation

Force

Read Explanation

Electrostatics

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