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Chapter 33: Q. 36 (page 956)

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" 4∆yin 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=0.2×10−3m×0.5×10−2m620×10−9mL=1.61m.

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Most popular questions from this chapter

A Michelson interferometer uses light from a sodium lamp. Sodium atoms emit light having wavelengths 589.0nmand 589.6nm. The interferometer is initially set up with both arms of equal length L1=L2, producing a bright spot at the center of the interference pattern. How far must mirrorM2be moved so that one wavelength has produced one more new maximum than the other wavelength?

Infrared light of wavelength 2.5μmilluminates a 0.20-mmdiameter hole. What is the angle of the first dark fringe in radians? In degrees?

One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a small patch of light on the far wall. Having recently studied optics in your physics class, you're not too surprised to see that the patch of light seems to be a circular diffraction pattern. It appears that the central maximum is about 1cmacross, and you estimate that the distance from the window shade to the wall is about 3m.. Estimate (a) the average wavelength of the sunlight (in nm ) and (b) the diameter of the pinhole (in mm ).

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.

FIGURE shows light of wavelength λincident at angle ϕon a reflection grating of spacing d. We want to find the angles um at which constructive interference occurs.

a. The figure shows paths 1and 2along which two waves travel and interfere. Find an expression for the path-length difference Δr=r2−r1.33

b. Using your result from part a, find an equation (analogous to Equation localid="1650299740348" (33.15)for the angles localid="1650299747450" θmat which diffraction occurs when the light is incident at angle localid="1650299754268" ∅. Notice that m can be a negative integer in your expression, indicating that path localid="1650299766020" 2is shorter than path localid="1650299773517" 1.

c. Show that the zeroth-order diffraction is simply a “reflection.” That is, localid="1650299781268" θ0=ϕ

d. Light of wavelength 500 nm is incident at localid="1650299787850" ϕ=40∘on a reflection grating having localid="1650299794954" 700reflection lines/mm. Find all angles localid="1650299802944" θmat which light is diffracted. Negative values of localid="1650299812949" θm
are interpreted as an angle left of the vertical.

e. Draw a picture showing a single localid="1650299823499" 500nmlight ray incident at localid="1650299833529" ϕ=40∘and showing all the diffracted waves at the correct angles.
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