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Chapter 33: Q.28 (page 956)

Your artist friend is designing an exhibit inspired by circular-aperture diffraction. A pinhole in a red zone is going to be illuminated with a red laser beam of wavelength 670nm, while a pinhole in a violet zone is going to be illuminated with a violet laser beam of wavelength 410nm. She wants all the diffraction patterns seen on a distant screen to have the same size. For this to work, what must be the ratio of the red pinhole’s diameter to that of the violet pinhole?

Short Answer

Expert verified

Ratio of red pinhole diameter to violet pinhole ,DredDviolet=1.63

Step by step solution

01

Laser Beam

A laser beam is a single-wavelength stream of concentrated, coherent light.

02

Find ratio

In order for the diffraction patterns from the two pinholes to have the same size, the angle of the first minimum in the two patterns must be the same, where this angle is given by the following relation

θ1=1.22λD

When you multiply θ1by the number of pinholes, you get

1.22λredDred=1.22λvioletDviolet

rearranging the equations,

DredDviolet=λredλviolet=670nm410nm=1.63

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Most popular questions from this chapter

FIGUREP33.36shows the light intensity on a screen behind a double slit. Suppose one slit is covered. What will be the light intensity at the center of the screen due to the remaining slit?

A Michelson interferometer uses red light with a wavelength of 656.45nm from a hydrogen discharge lamp. How many bright-dark-bright fringe shifts are observed if mirror M2 is moved exactly 1cm?

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.

Light of wavelength550nm illuminates a double slit, and the interference pattern is observed on a screen. At the position of the m=2 bright fringe, how much farther is it to the more distant slit than to the nearer slit?

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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