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

Short Answer

Expert verified

Angle of the first dark fringe in radians and in degree is 0.01525rador0.87

Step by step solution

01

Introduction

Dark Fringe:

Destructive interference causes the dark fringes to appear.

If the resultant amplitude and hence the resultant intensity are both zero, interference is said to be destructive.

02

Find angle of dark fringe 

The angle at which the first dark fringe appears in the circular aperture experiment is given by

1=1.22D

As a result, in our scenario, we'll have

localid="1650215776407" 1=1.222.510-6m210-4m

=0.01525rad

03

Dark fringe 

It's easy to convert to degrees if we remember that the complete angle 360equals 2. As a result, our perspective will be

localid="1650213775027" 1=0.01525rad3602

=0.87

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Most popular questions from this chapter

a. Green light shines through a 100-mm-diameter hole and is observed on a screen. If the hole diameter is increased by 20%, does the circular spot of light on the screen decrease in diameter, increase in diameter, or stay the same? Explain.

b. Green light shines through a 100m-diameter hole and is observed on a screen. If the hole diameter is increased by20%, does the circular spot of light on the screen decrease in diameter, increase in diameter, or stay the same? Explain.

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.

You've found an unlabeled diffraction grating. Before you can use it, you need to know how many lines per it has. To find out, you illuminate the grating with light of several different wavelengths and then measure the distance between the two first-order bright fringes on a viewing screen 150cmbehind the grating. Your data are as follows:


Use the best-fit line of an appropriate graph to determine the number of lines per mm.

Two 50-m-wide slits spaced 0.25mmapart are illuminated by blue laser light with a wavelength of 450nm. The interference pattern is observed on a screen2.0m behind the slits. How many bright fringes are seen in the central maximum that spans the distance between the first missing order on one side and the first missing order on the other side?

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.

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