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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 θ1=1.22λD, so:

As λis 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

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.

Light from a helium-neon laser (λ=633nm)passes through a circular aperture and is observed on a screen 4.0mbehind the aperture. The width of the central maximum is2.5cm. What is the diameter (in mm) of the hole?

A laser beam with a wavelength of 480nm illuminates two 0.12-mm-wide slits separated by 0.30mm. The interference pattern is observed on a screen 2.3m behind the slits. What is the light intensity, as a fraction of the maximum intensity I0, at a point halfway between the center and the first minimum?

A diffraction grating produces a first-order maximum at an angle of 20.0°. What is the angle of the second-order maximum?

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?

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