/*! 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} Q. 72 The intensity at the central max... [FREE SOLUTION] | 91影视

91影视

The intensity at the central maximum of a double-slit interference pattern is 4I1. The intensity at the first minimum is zero. At what fraction of the distance from the central maximum to the first minimum is the intensity I1? Assume an ideal double slit.

Short Answer

Expert verified

The intensity is the fraction of the distance between the highest and the minimum. yy0=23

Step by step solution

01

Introduction

The average power transmission across one period of a wave, such as acoustic waves (sound) or electromagnetic waves like light or radio waves, is referred to as intensity. Intensity can be used to portray intensity in a variety of contexts.

02

Find the fraction

The intensity of a perfect double slit can be calculated using the equation below.

Idouble=4I1cos2dLy

Let's substitute Itext double with I1obtain the ratio of the width between the centre high and the first minimum at which the intensity becomes I1.

Idouble4I1=I14I1=cos2dLy

cos2dLy=14

The original number of both sides yields

cosdLy=12cos112=dLy

to separate y, change the equation

y=Ldcos112

This is the location where the intensity becomes I1, However, since our purpose is to find the ratio of this location to the first minimum position, we'll apply the phrase below to find the first minimum position.

ym=m+12Ldm=0,1,2,

As a result, that the very first minimum's position is

y0=L2d

As a result, the width fraction between the centre best and the first minimum when the intensities is increased. I1is

yy0=Ldcos1(0.5)L2d=23=0.666

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!

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

It shows the light intensity on a viewing screen behind a circular aperture. What happens to the width of the central maximum if the

a. The wavelength of the light is increased.

b. The diameter of the aperture is increased.

c. How will the screen appear if the aperture diameter is less than the light wavelength?

FIGURE Q33.5 shows the light intensity on a viewing screen behind a single slit of width a The light's wavelength is. Is<a,=a,>a, or is it not possible to tell? Explain

A diffraction grating with 600linesmmis illuminated with light of wavelength 510nm. A very wide viewing screen is2.0m behind the grating.
aWhat is the distance between the twom=1 bright fringes?
bHow many bright fringes can be seen on the screen?

Light from a helium-neon laser (=633nm)illuminates a circular aperture. It is noted that the diameter of the central maximum on a screen 50cmbehind the aperture matches the diameter of the geometric image. What is the aperture's diameter (in mm)?

FIGURE shows light of wavelength incident at angle on a reflection grating of spacing d. We want to find the angles um at which constructive interference occurs.

a. The figure shows paths 1and 2along which two waves travel and interfere. Find an expression for the path-length difference r=r2r1.33

b. Using your result from part a, find an equation (analogous to Equation localid="1650299740348" (33.15)for the angles localid="1650299747450" mat which diffraction occurs when the light is incident at angle localid="1650299754268" . Notice that m can be a negative integer in your expression, indicating that path localid="1650299766020" 2is shorter than path localid="1650299773517" 1.

c. Show that the zeroth-order diffraction is simply a 鈥渞eflection.鈥 That is, localid="1650299781268" 0=

d. Light of wavelength 500 nm is incident at localid="1650299787850" =40on a reflection grating having localid="1650299794954" 700reflection lines/mm. Find all angles localid="1650299802944" mat which light is diffracted. Negative values of localid="1650299812949" m
are interpreted as an angle left of the vertical.

e. Draw a picture showing a single localid="1650299823499" 500nmlight ray incident at localid="1650299833529" =40and showing all the diffracted waves at the correct angles.

See all solutions

Recommended explanations on Physics Textbooks

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.