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Chapter 33: Q. 26 (page 956)

Infrared light of wavelength 2.5μmilluminates 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

θ1=1.22λD

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

localid="1650215776407" θ1=1.22×2.5×10-6m2×10-4m

=0.01525rad

03

Dark fringe 

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

localid="1650213775027" θ1=0.01525rad×3602π

=0.87°

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Most popular questions from this chapter

A 0.50-mm-diameter hole is illuminated by light of wavelength 550nm. What is the width (in mm) of the central maximum on a screen 2.0m behind the slit?

Light of wavelength620nmilluminates a diffraction grating. The second-order maximum is at angle 39.5°. How many lines per millimeter does this grating have?

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 wings of some beetles have closely spaced parallel lines of melanin, causing the wing to act as a reflection grating. Suppose sunlight shines straight onto a beetle wing. If the melanin lines on the wing are spaced 2.0μmapart, what is the first-order diffraction angle for green light λ=550nm?

aFind an expression for the positions y1of the first-order fringes of a diffraction grating if the line spacing is large enough for the small-angle approximation tan⁡θ≈sin⁡θ≈θto be valid. Your expression should be in terms of d,Landλ.
b. Use your expression from part a to find an expression for the separation∆yon the screen of two fringes that differ in wavelength by∆λ.
cRather than a viewing screen, modern spectrometers use detectors-similar to the one in your digital camera-that are divided into pixels. Consider a spectrometer with a 333lines/mmgrating and a detector with 100pixels/mmlocated 12cmbehind the grating. The resolution of a spectrometer is the smallest wavelength separation ∆λminthat can be measured reliably. What is the resolution of this spectrometer for wavelengths near localid="1649156925210" 550nm, in the center of the visible spectrum? You can assume that the fringe due to one specific wavelength is narrow enough to illuminate only one column of pixels.
See all solutions

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