Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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=4I1cos2πdλLy

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=cos2πdλLy

cos2πdλLy=14

The original number of both sides yields

cosπdλLy=12cos112=πdλLy

to separate y, change the equation

y=λLπdcos112

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+12λLdm=0,1,2,

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

y0=λL2d

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

yy0=λLπdcos1(0.5)λL2d=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 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

FIGUREP33.36shows the light intensity on a screen behind a double slit. The slit spacing is 0.20mmand the screen is 2.0mbehind the slits. What is the wavelength (in nm) of the light?

Light of wavelength550nm illuminates a double slit, and the interference pattern is observed on a screen. At the position of the m=2 bright fringe, how much farther is it to the more distant slit than to the nearer slit?

The pinhole camera of FIGURE images distant objects by allowing only a narrow bundle of light rays to pass through the hole and strike the film. If light consisted of particles, you could make the image sharper and sharper (at the expense of getting dimmer and dimmer) by making the aperture smaller and smaller. In practice, diffraction of light by the circular aperture limits the maximum sharpness that can be obtained. Consider two distant points of light, such as two distant streetlights. Each will produce a circular diffraction pattern on the film. The two images can just barely be resolved if the central maximum of one image falls on the first dark fringe of the other image. (This is called Rayleigh’s criterion, and we will explore its implication for optical instruments in Chapter 35.)

a. Optimum sharpness of one image occurs when the diameter of the central maximum equals the diameter of the pinhole. What is the optimum hole size for a pinhole camera in which the film is 20cmbehind the hole? Assume localid="1649089848422" λ=550nman average value for visible light.

b. For this hole size, what is the angle a (in degrees) between two distant sources that can barely be resolved?

c. What is the distance between two street lights localid="1649089839579" 1kmaway that can barely be resolved?

The two most prominent wavelengths in the light emitted by a hydrogen discharge lamp are 656nm (red) and 486nm (blue). Light from a hydrogen lamp illuminates a diffraction grating with 500lines/mm, and the light is observed on a screen 1.50m behind the grating. What is the distance between the first-order red and blue fringes?

Infrared light of wavelength 2.5μmilluminates a 0.20-mmdiameter hole. What is the angle of the first dark fringe in radians? In degrees?

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.

Sign-up for free