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 38.72AP

(a)

To determine

To show: The point where I=0.5Imax must have ϕ=2sinϕ .

(a)

Expert Solution
Check Mark

Answer to Problem 38.72AP

The point where I=0.5Imax must have ϕ=2sinϕ .

Explanation of Solution

Given info: The equation of the intensity of the light in the diffraction pattern is I=Imaxsin2ϕϕ2 where ϕ=(πasinθ)λ

The formula to calculate the intensity of the light is,

I=Imaxsin2ϕϕ2IImax=sin2ϕϕ2

Here,

Imax is the maximum intensity of the light.

ϕ is the phase constant of the light.

The value of I=0.5Imax .

Substitute 0.5Imax for I in above equation to find the value of ϕ .

0.5ImaxImax=sin2ϕϕ2sin2ϕϕ2=12sinϕ=ϕ2ϕ=2sinϕ

Conclusion

Therefore, the point where I=0.5Imax must have ϕ=2sinϕ .

(b)

To determine

To draw: Plot y1=sinϕ and y2=ϕ2 on the same set of axes over a range from ϕ=1rad to ϕ=π2rad .

(b)

Expert Solution
Check Mark

Answer to Problem 38.72AP

The graph between y1=sinϕ and y2=ϕ2 on the same set of axes over a range from ϕ=1rad to ϕ=π2rad .

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 38, Problem 38.72AP , additional homework tip  1

Figure (1)

Explanation of Solution

Given info: The equation of the intensity of the light in the diffraction pattern is I=Imaxsin2ϕϕ2 where ϕ=(πasinθ)λ

The equation of y1 is sinϕ and the equation for y2 is ϕ2 over a range from ϕ=1rad to ϕ=π2rad is shown in the figure below.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 38, Problem 38.72AP , additional homework tip  2

The solution of both the equation to coincide at a point is ,

sinϕ=ϕ2ϕ=1.39rad

So the solution of the transcendental equation is ϕ=1.39rad .

(c)

To determine

To show: The angular full width at half maximum of the central diffraction maximum is θ=0.885λa if the fraction λa is not large.

(c)

Expert Solution
Check Mark

Answer to Problem 38.72AP

The angular full width at half maximum of the central diffraction maximum is θ=0.885λa if the fraction λa is not large.

Explanation of Solution

Given info: The equation of the intensity of the light in the diffraction pattern is I=Imaxsin2ϕϕ2 where ϕ=(πasinθ)λ

The formula to calculate the phase angle is,

ϕ=(πasinθ)λ

Rewrite the above equation for sinθ .

sinθ=(ϕπ)λa

If the value of λa is small then,

θ(ϕπ)λa

The path covered by the light is symmetric so the phase angle is double the initial value.

θ=2(ϕπ)λa

Substitute 1.39rad for ϕ in above equation to find the value of θ .

θ=2(1.39rad3.14rad)λa=0.885λa

Conclusion

Therefore, the angular full width at half maximum of the central diffraction maximum is θ=0.885λa if the fraction λa is not large.

(d)

To determine

The number of steps involved to solve the transcendental equation ϕ=2sinϕ .

(d)

Expert Solution
Check Mark

Answer to Problem 38.72AP

The number of steps involved to solve the transcendental equation ϕ=2sinϕ is around 13 .

Explanation of Solution

Given info: The equation of the intensity of the light in the diffraction pattern is I=Imaxsin2ϕϕ2 where ϕ=(πasinθ)λ

The equation of y1 is 2sinϕ and the equation for y2 is ϕ , the value of ϕ is taken from 1 to 2 to find the solution of the equation and the corresponding values are shown in the table below.

ϕ 2sinϕ
1 1.19
2 1.29
1.5 1.41
1.4 1.394
1.39 1.391
1.392 1.3917
1.3915 1.39154
1.39152 1.39155
1.3916 1.39158
1.39158 1.391563
1.39157 1.391561
1.39156 1.391558
1.3915574 1.3915574

The solution of the transcendental equation ϕ=2sinϕ is 1.3915574 and the solution can be achieved in around 13 steps.

Conclusion

Therefore, the number of steps involved to solve the transcendental equation ϕ=2sinϕ is around 13 .

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How much diffraction spreading does a light beam undergo? One quantitative answer is the full width at half maximum of the central maximum of the single-slit Fraunhofer diffraction pattern. You can evaluate this angle of spreading in this problem. (a) as shown, define φ = πa sin φ/λ and show that at the point where I = 0.5Imax we must have φ = √2 sin φ. (b) Let y1 = sin φ and y2 = φ = /√2. Plot y1 and y2 on the same set of axes over a range from φ = 1 rad to φ = π/2 rad. Determine φ from the point of intersection of the two curves. (c) Then show that if the fraction λ/a is notlarge, the angular full width at half maximum of the central diffraction maximum is θ = 0.885λ/a. (d) What If? Another method to solve the transcendental equation φ = √2 sin φ in part (a) is to guess a first value of φ, use a computer or calculator to see how nearly it fits, and continue to update your estimate until the equation balances. How many steps(iterations) does this process take?
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Chapter 38 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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