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Chapter 33: Q. 8 (page 954)

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 100−μm-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.

Short Answer

Expert verified

(a) The rounded spot of light display on the screen increases.

(b) The rounded spot of light display on the screen decreses.

Step by step solution

01

Introduction

A hole diameter, often known as a PHD, is the size of a producing tool used to drill holes. As a result, the hole size diameter estimate for non-plated through holes and plated through holes differed.

02

Explanation

By increasing the hole diameter, the size of both the spot on the screen will grow. The reason for this is that when the diameter grows bigger, ray glasses will take precedence over wave optics.

03

Explanation

Increasing the diameter of the hole by 20%reduces the diameter of the spot on the screen. We must address diffraction effects when the hole is close to 1mmbecause it is of the same order as the wavelength of light.

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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?

a. Green light will shine through a hole with a 100-mm-diameter hole of and is seen on a screen. Does the circular point of lights 20%,is on the screen shrink in diameter, rise in diameter, or remain the same in diameter by? Explain.

b. Green light will shine through a hole with a 100-μmdiameter of and is seen on a screen. Does the circular point of lights is 20%,on the screen shrink in diameter, rise in diameter, or remain the same in diameter by? Explain.

Light of wavelength 600nmpasses though two slits separated by 0.20mmand is observed on a screen 1.0mbehind the slits. The location of the central maximum is marked on the screen and labeled y=0.

a. At what distance, on either side of y=0, are the m=1bright fringes?

b. A very thin piece of glass is then placed in one slit. Because light travels slower in glass than in air, the wave passing through the glass is delayed by 5.0×10-16sin comparison to the wave going through the other slit. What fraction of the period of the light wave is this delay?

c. With the glass in place, what is the phase difference Δϕ0between the two waves as they leave the slits?2

d. The glass causes the interference fringe pattern on the screen to shift sideways. Which way does the central maximum move (toward or away from the slit with the glass) and by how far?

FIGURE shows two nearly overlapped intensity peaks of the sort you might produce with a diffraction grating . As a practical matter, two peaks can just barely be resolved if their spacing ∆yequals the width w of each peak, where wis measured at half of the peak’s height. Two peaks closer together than wwill merge into a single peak. We can use this idea to understand the resolution of a diffraction grating.

a. In the small-angle approximation, the position of the m=1peak of a diffraction grating falls at the same location as the m=1fringe of a double slit: y1=λL/d. Suppose two wavelengths differing by ∆lpass through a grating at the same time. Find an expression for localid="1649086237242" ∆y, the separation of their first-order peaks.

b. We noted that the widths of the bright fringes are proportional to localid="1649086301255" 1/N, where localid="1649086311478" Nis the number of slits in the grating. Let’s hypothesize that the fringe width is localid="1649086321711" w=y1/NShow that this is true for the double-slit pattern. We’ll then assume it to be true as localid="1649086339026" Nincreases.

c. Use your results from parts a and b together with the idea that localid="1649086329574" Δymin=wto find an expression for localid="1649086347645" Δλmin, the minimum wavelength separation (in first order) for which the diffraction fringes can barely be resolved.

d. Ordinary hydrogen atoms emit red light with a wavelength of localid="1649086355936" 656.45nm.In deuterium, which is a “heavy” isotope of hydrogen, the wavelength is localid="1649086363764" 656.27nm.What is the minimum number of slits in a diffraction grating that can barely resolve these two wavelengths in the first-order diffraction pattern?

FIGURE P33.56 shows the light intensity on a screen behind a circular aperture. The wavelength of the light is 500nmand the screen is 1.0mbehind the slit. What is the diameter (in mm) of the aperture?
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