FIGUREP33.36shows the light intensity on a screen behind a double slit. The slit spacing is 0.20mmand the wavelength of the light is 620nm. What is the distance from the slits to the screen?

Short Answer

Expert verified

The screen's length from the slits isL=1.61m.

Step by step solution

01

Step: 1 Screen slit:

The fringe width reduced as the slit spacing improved, indicating reduced interference. The fringe width rose as the distance between the slits and the wall improved, because the light would have more area to diffract outwards and so interfere more.

02

Step: 2 Finding spacing fringes:

The distance between any two subsequent fringes in a double slit experiment is provided by the following relation.

Δy=λLd.

The spacing between five peaks is comparable to role="math" localid="1649086556897" 4yin FIGUREP33.36. As a result, the distance between any two consecutive fringes is

Δy=2cm4Δy=0.5cm.

03

Step: 3 Finding slit screen distance:

Distance of screen slit by

L=dΔyλL=0.2×103m×0.5×102m620×109mL=1.61m.

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Most popular questions from this chapter

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?

A Michelson interferometer is set up to display constructive interference (a bright central spot in the fringe pattern of Figure) using light of wavelength l. If the wavelength is changed to λ/2, does the central spot remain bright, does the central spot become dark, or do the fringes disappear? Explain. Assume the fringes are viewed by a detector sensitive to both wavelengths.

For your science fair project you need to design a diffraction grating that will disperse the visible spectrum 400-700nmover30.0 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 λ=589nm?

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.

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?

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