EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
9th Edition
ISBN: 9780100454897
Author: Jewett
Publisher: YUZU
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Chapter 38, Problem 38.77CP

Suppose the single slit in Figure 38.4 is 6.00 cm wide and in front of a microwave source operating at 7.50 GHz. (a) Calculate the angle for the first minimum in the diffraction pattern. (b) What is the relative intensity l/lmax at 0 =- 15.0°? (c) Assume two such sources, separated laterally by 20.0 cm, are behind the slit. What must be the maximum distance between the plane of the sources and the slit if the diffraction pat­terns are to be resolved? In this case, the approximation sin θ ≈ tan 8 is not valid because of the relatively small value of a/λ.

(a)

Expert Solution
Check Mark
To determine

The angle for the first minimum in the diffraction pattern.

Answer to Problem 38.77CP

The angle for the first minimum in the diffraction pattern is 41.8° .

Explanation of Solution

Given info: The width of the slit is 6.00cm and the microwave is operating at 7.50GHz .

The expression of wavelength ( λ ) of the microwave is,

λ=cf

Here,

c is the speed of light.

f is the frequency of microwave.

Substitute 7.50GHz for f and 3×108m/s in the above equation.

λ=3×108m/s7.50GHz(109Hz1GHz)=3×108m/s7.50×109Hzλ=0.04m

Thus, the wavelength of the microwave source is 0.04m .

The expression of the condition for the first minimum in the diffraction pattern is,

sinθ=mλa

Here,

m is the number of minimum.

λ is the wavelength of the microwave source.

a is the width of the slit.

Substitute 1 for m , 6.00cm for a and 0.04m for λ in the above equation.

sinθ=1(0.04m)6.00cm(102m1cm)sinθ=0.666θ=sin1(0.666)θ41.8°

Thus, the angle for the first minimum in the diffraction pattern is 41.8° .

Conclusion:

Therefore, the angle for the first minimum in the diffraction pattern is 41.8° .

(b)

Expert Solution
Check Mark
To determine

The relative intensity at 15.0° .

Answer to Problem 38.77CP

The relative intensity at 15.0° is 0.592 .

Explanation of Solution

Given info: The width of the slit is 6.00cm , the angle to find the relative intensity is 15.0° and the microwave is operating at 7.50GHz .

The expression of the intensity variation in a diffraction pattern from a single slit is,

I=Imax[sin(πasinθλ)πasinθλ]2

Here,

Imax is the intensity at central maximum.

λ is the wavelength of the microwave source.

Rearrange the above equation for IImax .

I=Imax[sin(πasinθλ)πasinθλ]2IImax=[sin(πasinθλ)πasinθλ]2

Substitute 15.0° for θ , 6.00cm for a and 0.04m for λ in the above equation.

IImax=[sin(π×6.00cm(102m1cm)sin15.0°0.04m)π×6.00cm(102m1cm)sin15.0°0.04m]2=[sin(π×0.06m(0.2588)0.04m)π×0.06m(0.2588)0.04m]2IImax=0.592

Thus, the relative intensity IImax at 15.0° is 0.592 .

Conclusion

Therefore, the relative intensity at 15.0° is 0.592 .

(c)

Expert Solution
Check Mark
To determine

The maximum distance between the plane of the sources and the slit if the diffraction pattern are to be resolved.

Answer to Problem 38.77CP

The maximum distance between the plane of the sources and the slit if the diffraction pattern are to be resolved is 0.262m .

Explanation of Solution

Given info: The width of the slit is 6.00cm , the microwave is operating at 7.50GHz and the distance between the two sources is 20.0cm .

The figure1 shows the given condition.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 38, Problem 38.77CP

Figure (1)

Consider β be the angle halfway between the two sources.

The expression of distance ( l ) of each source from the central line is,

l=D2

Here,

D is the distance between the two sources.

Substitute 20.0cm for D in the above equation.

l=20.0cm(102m1cm)2l=0.100m

Thus, the distance of each source from the central line is 0.100m .

From figure1 the expression of distance ( L ) between the plane of sources and the slit is,

L=lcotβ

The value of angle β is equal to angle θ2 according to vertically opposite angle.

Substitute θ2 for β in the above equation.

L=lcot(θ2)

Substitute 41.8° for θ and 0.100m for l in the above equation.

L=(0.100m)cot(41.8°2)=(0.100m)cot(20.9°)=0.2618m=0.262m

Conclusion:

Therefore, The maximum distance between the plane of the sources and the slit if the diffraction pattern are to be resolved is 0.262m .

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Chapter 38 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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