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Chapter 33: Q. 6 (page 954)

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

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?

Short Answer

Expert verified

(a) The width of the wavelength is increased.

(b) The width of the aperture is decreased.

(c) The screen appears like almost uniformly gray.

Step by step solution

01

Introduction

When it comes to waves like acoustic waves (sound) or electromagnetic waves like light or radio waves, intensity refers to the average power transfer across one period of the wave. Intensity can be used in a variety of situations where energy is delivered. For example, the intensity of the kinetic energy carried by drops of water from a garden sprinkler may be calculated.

02

Find the width (part a)

We know that 胃1=1.22位D, so:

As 位is increased, the width grows.

03

Find the width (part b)

As the diameter grows larger, the width shrinks.

04

Appearance of screen

For diffraction to occur, the field of view must be of the same order as the wavelength of light, resulting in a nearly evenly grey surface with no minima.

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Most popular questions from this chapter

For your science fair project you need to design a diffraction grating that will disperse the visible spectrum 400-700nmover30.0鈭� in first order.
a How many lines per millimeter does your grating need?
bWhat is the first-order diffraction angle of light from a sodium lamp 位=589nm?

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 鈭�yequals the width w of each peak, where wis measured at half of the peak鈥檚 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=位L/d. Suppose two wavelengths differing by 鈭�lpass through a grating at the same time. Find an expression for localid="1649086237242" 鈭�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鈥檚 hypothesize that the fringe width is localid="1649086321711" w=y1/NShow that this is true for the double-slit pattern. We鈥檒l 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" 螖ymin=wto find an expression for localid="1649086347645" 螖位min, 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 鈥渉eavy鈥� 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?

For what slit-width-to-wavelength ratio does the first minimum of a single-slit diffraction pattern appear at (a) 30掳, (b) 60掳, and (c) 90掳?

On a screen behind such a diffracting grating, narrow, bright fringes can be seen. After that, the entire research is submerged in water. Do the screen's fringes get closer together, farther apart, stay the same, or simply disappear? Explain

A diffraction grating having 500lines/mmdiffracts visible light at 30鈭�.What is the light's wavelength?
See all solutions

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