Physics for Scientists and Engineers with Modern Physics, Technology Update
Physics for Scientists and Engineers with Modern Physics, Technology Update
9th Edition
ISBN: 9781305401969
Author: SERWAY, Raymond A.; Jewett, John W.
Publisher: Cengage Learning
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Chapter 38, Problem 27P

Consider an array of parallel wires with uniform spacing of 1.30 cm between centers. In air at 20.0°C, ultrasound with a frequency of 37.2 kHz from a distant source is incident perpendicular to the array. (a) Find the number of directions on the other side of the array in which there is a maximum of intensity. (b) Find the angle for each of these directions relative to the direction of the incident beam.

(a)

Expert Solution
Check Mark
To determine

The number of directions on the other side of the array for maximum intensity.

Answer to Problem 27P

The number of directions on the other side of the array for maximum intensity is three.

Explanation of Solution

Given info: Temperature of air is 20.0°C , spacing between centre is 1.30cm and frequency of array is 37.2kHz .

The wavelength for a diffraction grating can be given as,

λ=vf

Here,

λ is the wavelength of light.

v is the speed of sound.

f is the frequency of the array.

Substitute 343m/s for v and 37.2kHz for f in the above equation to find λ ,

λ=(343m/s)(37.2kHz)(103Hz1kHz)=9.22×103m

The condition for the bright fringe in diffraction can be given as,

mλ=dsinθ (1)

Here,

θ is the angle of spectral line.

λ is the wavelength of light.

m is the order of diffraction.

d is the spacing between centre.

Substitute 90° for θ , 9.22×103m for λ and 1.30cm for d in the equation (1),

m(9.22×103m)=[(1.30cm)(1m100cm)]sin90°m=1.411

The maximum number of direction possible can be given as,

mmax=2m+1

Here,

mmax is the maximum number of directions.

Substitute 1 for m in the above equation,

mmax=2(1)+1=3

Thus, the number of directions on the other side of the array for maximum intensity is three.

Conclusion:

Therefore, the number of directions on the other side of the array for maximum intensity is three.

(b)

Expert Solution
Check Mark
To determine

The angle for each of the directions relative to the direction of the incident beam.

Answer to Problem 27P

The angle for each of the directions relative to the direction of the incident beam is 0° , +45.2° and 45.2° .

Explanation of Solution

Given info: Temperature of air is 20.0°C , spacing between centre is 1.30cm and frequency of array is 37.2kHz .

The condition for a diffraction grating as in equation (1) can be given as,

mλ=dsinθ

Rearrange the above expression for θ ,

θ=sin1(mλd) (2)

Substitute (2) for m , 9.22×103m for λ and 1.30cm for d in the equation (2),

θ=sin1((2)(9.22×103m)(1.30cm)(1m100cm))=sin1(1.418)

As the range of sine function is [1,1] , the value of sin1(1.418) is undetermined. Therefore, |m| cannot be more than unity.

Substitute (1) for m , 9.22×103m for λ and 1.30cm for d in the equation (2),

θ=sin1((1)(9.22×103m)(1.30cm)(1m100cm))=sin1(0.709)=45.2°

Thus, θ is (45.2°) for the (1) order of diffraction.

Substitute 0 for m , 9.22×103m for λ and 1.30cm for d in the equation (2),

θ=sin1((0)(9.22×103m)(1.30cm)(1m100cm))=sin1(0)=0°

Thus, θ is 0° for the 0 order of diffraction.

Substitute (1) for m , 9.22×103m for λ and 1.30cm for d in the equation (2),

θ=sin1((1)(9.22×103m)(1.30cm)(1m100cm))=sin1(0.709)=45.2°

Thus, θ is 45.2° for the (1) order of diffraction.

Conclusion:

Therefore, the angle for each of the directions relative to the direction of the incident beam is 0° , +45.2° and 45.2° .

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Chapter 38 Solutions

Physics for Scientists and Engineers with Modern Physics, Technology Update

Ch. 38 - Prob. 4OQCh. 38 - Prob. 5OQCh. 38 - Prob. 6OQCh. 38 - Prob. 7OQCh. 38 - Prob. 8OQCh. 38 - Prob. 9OQCh. 38 - Prob. 10OQCh. 38 - Prob. 11OQCh. 38 - Prob. 12OQCh. 38 - Prob. 1CQCh. 38 - Prob. 2CQCh. 38 - Prob. 3CQCh. 38 - Prob. 4CQCh. 38 - Prob. 5CQCh. 38 - Prob. 6CQCh. 38 - Prob. 7CQCh. 38 - Prob. 8CQCh. 38 - Prob. 9CQCh. 38 - Prob. 10CQCh. 38 - Prob. 11CQCh. 38 - Prob. 12CQCh. 38 - Prob. 1PCh. 38 - Prob. 2PCh. 38 - Prob. 3PCh. 38 - Prob. 4PCh. 38 - Prob. 5PCh. 38 - Prob. 6PCh. 38 - Prob. 7PCh. 38 - Prob. 8PCh. 38 - Prob. 9PCh. 38 - Prob. 10PCh. 38 - Prob. 11PCh. 38 - Coherent light of wavelength 501.5 nm is sent...Ch. 38 - Prob. 13PCh. 38 - Prob. 14PCh. 38 - Prob. 15PCh. 38 - Prob. 16PCh. 38 - Prob. 17PCh. 38 - Prob. 18PCh. 38 - What is the approximate size of the smallest...Ch. 38 - Prob. 20PCh. 38 - Prob. 21PCh. 38 - Prob. 22PCh. 38 - Prob. 23PCh. 38 - Prob. 24PCh. 38 - Prob. 25PCh. 38 - Prob. 26PCh. 38 - Consider an array of parallel wires with uniform...Ch. 38 - Prob. 28PCh. 38 - Prob. 29PCh. 38 - A grating with 250 grooves/mm is used with an...Ch. 38 - Prob. 31PCh. 38 - Prob. 32PCh. 38 - Light from an argon laser strikes a diffraction...Ch. 38 - Show that whenever white light is passed through a...Ch. 38 - Prob. 35PCh. 38 - Prob. 36PCh. 38 - Prob. 37PCh. 38 - Prob. 38PCh. 38 - Prob. 39PCh. 38 - Prob. 40PCh. 38 - Prob. 41PCh. 38 - Prob. 42PCh. 38 - Prob. 43PCh. 38 - Prob. 44PCh. 38 - Prob. 45PCh. 38 - Prob. 46PCh. 38 - Prob. 47PCh. 38 - Prob. 48PCh. 38 - Prob. 49PCh. 38 - Prob. 50PCh. 38 - Prob. 51PCh. 38 - Prob. 52PCh. 38 - Prob. 53APCh. 38 - Prob. 54APCh. 38 - Prob. 55APCh. 38 - Prob. 56APCh. 38 - Prob. 57APCh. 38 - Prob. 58APCh. 38 - Prob. 59APCh. 38 - Prob. 60APCh. 38 - Prob. 61APCh. 38 - Prob. 62APCh. 38 - Prob. 63APCh. 38 - Prob. 64APCh. 38 - Prob. 65APCh. 38 - Prob. 66APCh. 38 - Prob. 67APCh. 38 - Prob. 68APCh. 38 - Prob. 69APCh. 38 - Prob. 70APCh. 38 - Prob. 71APCh. 38 - Prob. 72APCh. 38 - Prob. 73APCh. 38 - Light of wavelength 632.8 nm illuminates a single...Ch. 38 - Prob. 75CPCh. 38 - Prob. 76CPCh. 38 - Prob. 77CPCh. 38 - Prob. 78CPCh. 38 - Prob. 79CP
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