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Chapter 27, Problem 45P

(a)

To determine

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

(a)

Expert Solution
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Answer to Problem 45P

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)

To determine

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

(b)

Expert Solution
Check Mark

Answer to Problem 45P

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 27 Solutions

Bundle: Principles of Physics: A Calculus-Based Text, 5th + WebAssign Printed Access Card for Serway/Jewett's Principles of Physics: A Calculus-Based Text, 5th Edition, Multi-Term

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