Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Q. 66 (page 486)

Engineers are testing a new thin-film coating whose index of refraction is less than that of glass. They deposit a 560-nm-thick layer on glass, then shine lasers on it. A red laser with a wavelength of 640 nm has no reflection at all, but a violet laser with a wavelength of 400 nm has a maximum reflection. How the coating behaves at other wavelengths is unknown. What is the coating’s index of refraction?

Short Answer

Expert verified

The coating's index of refraction is n= 1.429

Step by step solution

01

The concept of refraction 

The change in direction of a wave passing from one medium to another or from a gradual change in the medium is known as the refraction.

02

Explanation of the solution

Here, d=560nm

λred=640nm

λblue=400nm

Thus, the solution of the refraction is n=λeme2d

Thus, the ratio of the constructive interference is λCλD=2ndmc2ndmd-12400640-md-12mcmd-12mc=0.625

In order to find out the substitute of the mCby the random numbers of role="math" localid="1649153848269" mD. Thus, the substitute of mDis 3.

Thus, role="math" localid="1649153919628" 3-12mc=1.6mc=4

By using the coating's index of refraction in equation is n=400×42×560n=1.429

Centering isn=1.429

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

qasdff

M?

FIGURE EX17.7 shows a standing wave on a string that is oscillating at 100 Hz. a. How many antinodes will there be if the frequency is increased to 200 Hz? b. If the tension is increased by a factor of 4, at what frequency will the string continue to oscillate as a standing wave that looks like the one in the figure?

||| A water wave is called a deep-water wave if the water’s depth

is more than one-quarter of the wavelength. Unlike the waves

we’ve considered in this chapter, the speed of a deep-water wave

depends on its wavelength:

v = B

gl

2p

Longer wavelengths travel faster. Let’s apply this to standing waves.

Consider a diving pool that is 5.0 m deep and 10.0 m wide. Standing

water waves can set up across the width of the pool. Because

water sloshes up and down at the sides of the pool, the boundary

conditions require antinodes at x = 0 and x = L. Thus a standing

water wave resembles a standing sound wave in an open-open tube.

a. What are the wavelengths of the first three standing-wave

modes for water in the pool? Do they satisfy the condition for

being deep-water waves?

b. What are the wave speeds for each of these waves?

c. Derive a general expression for the frequencies fm of the possible

standing waves. Your expression should be in terms of m, g, and L.

d. What are the oscillation periods of the first three standing wave

Two loudspeakers emit sound waves along the x-axis. A listener in front of both speakers hears a maximum sound intensity when speaker 2 is at the origin and speaker 1 is at x = 0.50 m. If speaker 1 is slowly moved forward, the sound intensity decreases and then increases, reaching another maximum when speaker 1 is at x = 0.90 m.

a. What is the frequency of the sound? Assume vsound = 340 m/s.

b. What is the phase difference between the speakers?
See all solutions

Recommended explanations on Physics Textbooks

Astrophysics

Read Explanation

Electromagnetism

Read Explanation

Fields in Physics

Read Explanation

Famous Physicists

Read Explanation

Fluids

Read Explanation

Classical Mechanics

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