In Figure P36.10 (not to scale), let L = 1.20 m and d = 0.120 mm and assume the slit system is illuminated with monochromatic 500-nm light. Calculate the phase difference between the two wave fronts arriving at P when (a) θ = 0.500° and (b) y = 5.00 mm. (c) What is the value of θ for which the phase difference is 0.333 rad? (d) What is the value of θ for which the path difference is λ /4? Figure P36.10

Question

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Answer 1

Textbook Question

Chapter 37, Problem 37.18P

In Figure P36.10 (not to scale), let L = 1.20 m and d = 0.120 mm and assume the slit system is illuminated with monochromatic 500-nm light. Calculate the phase difference between the two wave fronts arriving at P when (a) θ = 0.500° and (b) y = 5.00 mm. (c) What is the value of θ for which the phase difference is 0.333 rad? (d) What is the value of θ for which the path difference is λ/4?

Figure P36.10

Chapter 37, Problem 37.18P, In Figure P36.10 (not to scale), let L = 1.20 m and d = 0.120 mm and assume the slit system is

(a)

Expert Solution

To determine

The phase difference between the two waves fronts arriving at

$P$ when

$θ=0.500°$ .

Answer to Problem 37.18P

The phase difference between the two waves fronts arriving at

$P$ when

$θ=0.500°$ is

$13.2 radian$

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $500 nm$ .

The given diagram is shown below.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 37, Problem 37.18P

Figure 1

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

Substitute $0.120 mm$ for $d$ , $0.500°$ for $θ$ , $500 nm$ for $λ$ , in the above formula as,

$ϕ=2πλdsin θ=2π(500 nm×10−9 m1 nm)(.120 mm×10−3 m1 mm)sin (0.500°)=13.2 radian$

Conclusion:

Therefore, the phase difference between the two waves fronts arriving at $P$ when $θ=0.500°$ is $13.2 radian$

(b)

Expert Solution

To determine

The phase difference between the two waves fronts arriving at

$P$ when

$y=5.00 mm$ .

Answer to Problem 37.18P

The phase difference between the two waves fronts arriving at

$P$ when

$y=5.00 mm$ is

$6.28 radian$ .

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $6.28 radian$ .

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

From the right angle triangle $sin θ$ is

$sin θ=perpendicular hypotenuse=yL$

Here,

$y$ is the distance from $O$ to $P$ .

$L$ is the distance between slit and screen.

Substitute $yL$ for $sin θ$ in the above formula as,

$ϕ=2πλdsin θ=2πλd(yL)$

Substitute $0.120 mm$ for $d$ , $500 nm$ for $λ$ , $1.20 m$ for $L$ , $5.00 mm$ for $y$ in the above formula as,

$ϕ=2πλdsin θ=2π(500 nm×10−9 m1 nm)(.120 mm×10−3 m1 mm)(5.00 mm×10-3 m1 mm1.20 m)=6.28 radian$

Conclusion:

Therefore, the phase difference between the two waves fronts arriving at $P$ when $y=5.00 mm$ is $6.28 radian$

(c)

Expert Solution

To determine

The value of

$θ$ for which the path difference is

$λ4$ .

Answer to Problem 37.18P

The value of

$θ$ for which the path difference is

$λ4$ is

$1.27×10−2 degree$ .

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $6.28 radian$ , phase difference is $0.33 radian$

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

Rearrange the above formula to find $θ$ is,

$ϕ=2πλdsin θsin θ=λϕ2πdθ=sin−1 (λϕ2πd)$

Substitute $0.120 mm$ for $d$ , $500 nm$ for $λ$ , $0.33 radian$ for $ϕ$ in the above formula as,

$θ=sin−1 (λϕ2πd)=sin−1 (500 nm×10−9 mm1 mm(0.33 radian)2π(0.120 mm×10−3 m1 mm))=1.27×10−2 degree$

Conclusion:

Therefore, the value of $θ$ is $1.27×10−2 degree$ .

(d)

Expert Solution

To determine

The value of

$θ$

Answer to Problem 37.18P

The value of

$θ$ is

$5.97×10−2 degree$ .

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $6.28 radian$ , path difference is $λ4$

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$ (1)

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

The path difference is $λ4$ as,

$dsin θ=λ4$

Substitute $λ4$ for $dsin θ$ in the above formula as,

$ϕ=2πλdsin θ=2πλ(λ4)=π2$

Rearrange the equation (1) to find $θ$ as,

$θ=sin−1 (λϕ2πd)$

Substitute $0.120 mm$ for $d$ , $500 nm$ for $λ$ , $π2$ for $ϕ$ in the above formula as,

$θ=sin−1 (λ(π2)2πd)=sin−1 (λ4d)=sin−1 (500 nm×10−9 mm1 mm4(0.120 mm×10−3 m1 mm))=5.97×10−2 degree$

Conclusion:

Therefore, the value of $θ$ is. $5.97×10−2 degree$ .

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Answer 2

Textbook Question

Chapter 37, Problem 37.18P

In Figure P36.10 (not to scale), let L = 1.20 m and d = 0.120 mm and assume the slit system is illuminated with monochromatic 500-nm light. Calculate the phase difference between the two wave fronts arriving at P when (a) θ = 0.500° and (b) y = 5.00 mm. (c) What is the value of θ for which the phase difference is 0.333 rad? (d) What is the value of θ for which the path difference is λ/4?

Figure P36.10

Chapter 37, Problem 37.18P, In Figure P36.10 (not to scale), let L = 1.20 m and d = 0.120 mm and assume the slit system is

(a)

Expert Solution

To determine

The phase difference between the two waves fronts arriving at

$P$ when

$θ=0.500°$ .

Answer to Problem 37.18P

The phase difference between the two waves fronts arriving at

$P$ when

$θ=0.500°$ is

$13.2 radian$

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $500 nm$ .

The given diagram is shown below.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 37, Problem 37.18P

Figure 1

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

Substitute $0.120 mm$ for $d$ , $0.500°$ for $θ$ , $500 nm$ for $λ$ , in the above formula as,

$ϕ=2πλdsin θ=2π(500 nm×10−9 m1 nm)(.120 mm×10−3 m1 mm)sin (0.500°)=13.2 radian$

Conclusion:

Therefore, the phase difference between the two waves fronts arriving at $P$ when $θ=0.500°$ is $13.2 radian$

(b)

Expert Solution

To determine

The phase difference between the two waves fronts arriving at

$P$ when

$y=5.00 mm$ .

Answer to Problem 37.18P

The phase difference between the two waves fronts arriving at

$P$ when

$y=5.00 mm$ is

$6.28 radian$ .

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $6.28 radian$ .

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

From the right angle triangle $sin θ$ is

$sin θ=perpendicular hypotenuse=yL$

Here,

$y$ is the distance from $O$ to $P$ .

$L$ is the distance between slit and screen.

Substitute $yL$ for $sin θ$ in the above formula as,

$ϕ=2πλdsin θ=2πλd(yL)$

Substitute $0.120 mm$ for $d$ , $500 nm$ for $λ$ , $1.20 m$ for $L$ , $5.00 mm$ for $y$ in the above formula as,

$ϕ=2πλdsin θ=2π(500 nm×10−9 m1 nm)(.120 mm×10−3 m1 mm)(5.00 mm×10-3 m1 mm1.20 m)=6.28 radian$

Conclusion:

Therefore, the phase difference between the two waves fronts arriving at $P$ when $y=5.00 mm$ is $6.28 radian$

(c)

Expert Solution

To determine

The value of

$θ$ for which the path difference is

$λ4$ .

Answer to Problem 37.18P

The value of

$θ$ for which the path difference is

$λ4$ is

$1.27×10−2 degree$ .

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $6.28 radian$ , phase difference is $0.33 radian$

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

Rearrange the above formula to find $θ$ is,

$ϕ=2πλdsin θsin θ=λϕ2πdθ=sin−1 (λϕ2πd)$

Substitute $0.120 mm$ for $d$ , $500 nm$ for $λ$ , $0.33 radian$ for $ϕ$ in the above formula as,

$θ=sin−1 (λϕ2πd)=sin−1 (500 nm×10−9 mm1 mm(0.33 radian)2π(0.120 mm×10−3 m1 mm))=1.27×10−2 degree$

Conclusion:

Therefore, the value of $θ$ is $1.27×10−2 degree$ .

(d)

Expert Solution

To determine

The value of

$θ$

Answer to Problem 37.18P

The value of

$θ$ is

$5.97×10−2 degree$ .

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $6.28 radian$ , path difference is $λ4$

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$ (1)

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

The path difference is $λ4$ as,

$dsin θ=λ4$

Substitute $λ4$ for $dsin θ$ in the above formula as,

$ϕ=2πλdsin θ=2πλ(λ4)=π2$

Rearrange the equation (1) to find $θ$ as,

$θ=sin−1 (λϕ2πd)$

Substitute $0.120 mm$ for $d$ , $500 nm$ for $λ$ , $π2$ for $ϕ$ in the above formula as,

$θ=sin−1 (λ(π2)2πd)=sin−1 (λ4d)=sin−1 (500 nm×10−9 mm1 mm4(0.120 mm×10−3 m1 mm))=5.97×10−2 degree$

Conclusion:

Therefore, the value of $θ$ is. $5.97×10−2 degree$ .

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Answer 3

Textbook Question

Chapter 37, Problem 37.18P

In Figure P36.10 (not to scale), let L = 1.20 m and d = 0.120 mm and assume the slit system is illuminated with monochromatic 500-nm light. Calculate the phase difference between the two wave fronts arriving at P when (a) θ = 0.500° and (b) y = 5.00 mm. (c) What is the value of θ for which the phase difference is 0.333 rad? (d) What is the value of θ for which the path difference is λ/4?

Figure P36.10

Chapter 37, Problem 37.18P, In Figure P36.10 (not to scale), let L = 1.20 m and d = 0.120 mm and assume the slit system is

(a)

Expert Solution

To determine

The phase difference between the two waves fronts arriving at

$P$ when

$θ=0.500°$ .

Answer to Problem 37.18P

The phase difference between the two waves fronts arriving at

$P$ when

$θ=0.500°$ is

$13.2 radian$

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $500 nm$ .

The given diagram is shown below.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 37, Problem 37.18P

Figure 1

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

Substitute $0.120 mm$ for $d$ , $0.500°$ for $θ$ , $500 nm$ for $λ$ , in the above formula as,

$ϕ=2πλdsin θ=2π(500 nm×10−9 m1 nm)(.120 mm×10−3 m1 mm)sin (0.500°)=13.2 radian$

Conclusion:

Therefore, the phase difference between the two waves fronts arriving at $P$ when $θ=0.500°$ is $13.2 radian$

(b)

Expert Solution

To determine

The phase difference between the two waves fronts arriving at

$P$ when

$y=5.00 mm$ .

Answer to Problem 37.18P

The phase difference between the two waves fronts arriving at

$P$ when

$y=5.00 mm$ is

$6.28 radian$ .

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $6.28 radian$ .

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

From the right angle triangle $sin θ$ is

$sin θ=perpendicular hypotenuse=yL$

Here,

$y$ is the distance from $O$ to $P$ .

$L$ is the distance between slit and screen.

Substitute $yL$ for $sin θ$ in the above formula as,

$ϕ=2πλdsin θ=2πλd(yL)$

Substitute $0.120 mm$ for $d$ , $500 nm$ for $λ$ , $1.20 m$ for $L$ , $5.00 mm$ for $y$ in the above formula as,

$ϕ=2πλdsin θ=2π(500 nm×10−9 m1 nm)(.120 mm×10−3 m1 mm)(5.00 mm×10-3 m1 mm1.20 m)=6.28 radian$

Conclusion:

Therefore, the phase difference between the two waves fronts arriving at $P$ when $y=5.00 mm$ is $6.28 radian$

(c)

Expert Solution

To determine

The value of

$θ$ for which the path difference is

$λ4$ .

Answer to Problem 37.18P

The value of

$θ$ for which the path difference is

$λ4$ is

$1.27×10−2 degree$ .

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $6.28 radian$ , phase difference is $0.33 radian$

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

Rearrange the above formula to find $θ$ is,

$ϕ=2πλdsin θsin θ=λϕ2πdθ=sin−1 (λϕ2πd)$

Substitute $0.120 mm$ for $d$ , $500 nm$ for $λ$ , $0.33 radian$ for $ϕ$ in the above formula as,

$θ=sin−1 (λϕ2πd)=sin−1 (500 nm×10−9 mm1 mm(0.33 radian)2π(0.120 mm×10−3 m1 mm))=1.27×10−2 degree$

Conclusion:

Therefore, the value of $θ$ is $1.27×10−2 degree$ .

(d)

Expert Solution

To determine

The value of

$θ$

Answer to Problem 37.18P

The value of

$θ$ is

$5.97×10−2 degree$ .

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $6.28 radian$ , path difference is $λ4$

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$ (1)

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

The path difference is $λ4$ as,

$dsin θ=λ4$

Substitute $λ4$ for $dsin θ$ in the above formula as,

$ϕ=2πλdsin θ=2πλ(λ4)=π2$

Rearrange the equation (1) to find $θ$ as,

$θ=sin−1 (λϕ2πd)$

Substitute $0.120 mm$ for $d$ , $500 nm$ for $λ$ , $π2$ for $ϕ$ in the above formula as,

$θ=sin−1 (λ(π2)2πd)=sin−1 (λ4d)=sin−1 (500 nm×10−9 mm1 mm4(0.120 mm×10−3 m1 mm))=5.97×10−2 degree$

Conclusion:

Therefore, the value of $θ$ is. $5.97×10−2 degree$ .

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Answer 4

Textbook Question

Chapter 37, Problem 37.18P

In Figure P36.10 (not to scale), let L = 1.20 m and d = 0.120 mm and assume the slit system is illuminated with monochromatic 500-nm light. Calculate the phase difference between the two wave fronts arriving at P when (a) θ = 0.500° and (b) y = 5.00 mm. (c) What is the value of θ for which the phase difference is 0.333 rad? (d) What is the value of θ for which the path difference is λ/4?

Figure P36.10

Chapter 37, Problem 37.18P, In Figure P36.10 (not to scale), let L = 1.20 m and d = 0.120 mm and assume the slit system is

(a)

Expert Solution

To determine

The phase difference between the two waves fronts arriving at

$P$ when

$θ=0.500°$ .

Answer to Problem 37.18P

The phase difference between the two waves fronts arriving at

$P$ when

$θ=0.500°$ is

$13.2 radian$

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $500 nm$ .

The given diagram is shown below.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 37, Problem 37.18P

Figure 1

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

Substitute $0.120 mm$ for $d$ , $0.500°$ for $θ$ , $500 nm$ for $λ$ , in the above formula as,

$ϕ=2πλdsin θ=2π(500 nm×10−9 m1 nm)(.120 mm×10−3 m1 mm)sin (0.500°)=13.2 radian$

Conclusion:

Therefore, the phase difference between the two waves fronts arriving at $P$ when $θ=0.500°$ is $13.2 radian$

(b)

Expert Solution

To determine

The phase difference between the two waves fronts arriving at

$P$ when

$y=5.00 mm$ .

Answer to Problem 37.18P

The phase difference between the two waves fronts arriving at

$P$ when

$y=5.00 mm$ is

$6.28 radian$ .

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $6.28 radian$ .

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

From the right angle triangle $sin θ$ is

$sin θ=perpendicular hypotenuse=yL$

Here,

$y$ is the distance from $O$ to $P$ .

$L$ is the distance between slit and screen.

Substitute $yL$ for $sin θ$ in the above formula as,

$ϕ=2πλdsin θ=2πλd(yL)$

Substitute $0.120 mm$ for $d$ , $500 nm$ for $λ$ , $1.20 m$ for $L$ , $5.00 mm$ for $y$ in the above formula as,

$ϕ=2πλdsin θ=2π(500 nm×10−9 m1 nm)(.120 mm×10−3 m1 mm)(5.00 mm×10-3 m1 mm1.20 m)=6.28 radian$

Conclusion:

Therefore, the phase difference between the two waves fronts arriving at $P$ when $y=5.00 mm$ is $6.28 radian$

(c)

Expert Solution

To determine

The value of

$θ$ for which the path difference is

$λ4$ .

Answer to Problem 37.18P

The value of

$θ$ for which the path difference is

$λ4$ is

$1.27×10−2 degree$ .

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $6.28 radian$ , phase difference is $0.33 radian$

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

Rearrange the above formula to find $θ$ is,

$ϕ=2πλdsin θsin θ=λϕ2πdθ=sin−1 (λϕ2πd)$

Substitute $0.120 mm$ for $d$ , $500 nm$ for $λ$ , $0.33 radian$ for $ϕ$ in the above formula as,

$θ=sin−1 (λϕ2πd)=sin−1 (500 nm×10−9 mm1 mm(0.33 radian)2π(0.120 mm×10−3 m1 mm))=1.27×10−2 degree$

Conclusion:

Therefore, the value of $θ$ is $1.27×10−2 degree$ .

(d)

Expert Solution

To determine

The value of

$θ$

Answer to Problem 37.18P

The value of

$θ$ is

$5.97×10−2 degree$ .

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $6.28 radian$ , path difference is $λ4$

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$ (1)

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

The path difference is $λ4$ as,

$dsin θ=λ4$

Substitute $λ4$ for $dsin θ$ in the above formula as,

$ϕ=2πλdsin θ=2πλ(λ4)=π2$

Rearrange the equation (1) to find $θ$ as,

$θ=sin−1 (λϕ2πd)$

Substitute $0.120 mm$ for $d$ , $500 nm$ for $λ$ , $π2$ for $ϕ$ in the above formula as,

$θ=sin−1 (λ(π2)2πd)=sin−1 (λ4d)=sin−1 (500 nm×10−9 mm1 mm4(0.120 mm×10−3 m1 mm))=5.97×10−2 degree$

Conclusion:

Therefore, the value of $θ$ is. $5.97×10−2 degree$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The phase difference between the two waves fronts arriving at

$P$ when

$θ=0.500°$ .

Answer to Problem 37.18P

The phase difference between the two waves fronts arriving at

$P$ when

$θ=0.500°$ is

$13.2 radian$

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $500 nm$ .

The given diagram is shown below.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 37, Problem 37.18P

Figure 1

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

Substitute $0.120 mm$ for $d$ , $0.500°$ for $θ$ , $500 nm$ for $λ$ , in the above formula as,

$ϕ=2πλdsin θ=2π(500 nm×10−9 m1 nm)(.120 mm×10−3 m1 mm)sin (0.500°)=13.2 radian$

Conclusion:

Therefore, the phase difference between the two waves fronts arriving at $P$ when $θ=0.500°$ is $13.2 radian$

(b)

Expert Solution

To determine

The phase difference between the two waves fronts arriving at

$P$ when

$y=5.00 mm$ .

Answer to Problem 37.18P

The phase difference between the two waves fronts arriving at

$P$ when

$y=5.00 mm$ is

$6.28 radian$ .

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $6.28 radian$ .

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

From the right angle triangle $sin θ$ is

$sin θ=perpendicular hypotenuse=yL$

Here,

$y$ is the distance from $O$ to $P$ .

$L$ is the distance between slit and screen.

Substitute $yL$ for $sin θ$ in the above formula as,

$ϕ=2πλdsin θ=2πλd(yL)$

Substitute $0.120 mm$ for $d$ , $500 nm$ for $λ$ , $1.20 m$ for $L$ , $5.00 mm$ for $y$ in the above formula as,

$ϕ=2πλdsin θ=2π(500 nm×10−9 m1 nm)(.120 mm×10−3 m1 mm)(5.00 mm×10-3 m1 mm1.20 m)=6.28 radian$

Conclusion:

Therefore, the phase difference between the two waves fronts arriving at $P$ when $y=5.00 mm$ is $6.28 radian$

(c)

Expert Solution

To determine

The value of

$θ$ for which the path difference is

$λ4$ .

Answer to Problem 37.18P

The value of

$θ$ for which the path difference is

$λ4$ is

$1.27×10−2 degree$ .

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $6.28 radian$ , phase difference is $0.33 radian$

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

Rearrange the above formula to find $θ$ is,

$ϕ=2πλdsin θsin θ=λϕ2πdθ=sin−1 (λϕ2πd)$

Substitute $0.120 mm$ for $d$ , $500 nm$ for $λ$ , $0.33 radian$ for $ϕ$ in the above formula as,

$θ=sin−1 (λϕ2πd)=sin−1 (500 nm×10−9 mm1 mm(0.33 radian)2π(0.120 mm×10−3 m1 mm))=1.27×10−2 degree$

Conclusion:

Therefore, the value of $θ$ is $1.27×10−2 degree$ .

(d)

Expert Solution

To determine

The value of

$θ$

Answer to Problem 37.18P

The value of

$θ$ is

$5.97×10−2 degree$ .

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $6.28 radian$ , path difference is $λ4$

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$ (1)

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

The path difference is $λ4$ as,

$dsin θ=λ4$

Substitute $λ4$ for $dsin θ$ in the above formula as,

$ϕ=2πλdsin θ=2πλ(λ4)=π2$

Rearrange the equation (1) to find $θ$ as,

$θ=sin−1 (λϕ2πd)$

Substitute $0.120 mm$ for $d$ , $500 nm$ for $λ$ , $π2$ for $ϕ$ in the above formula as,

$θ=sin−1 (λ(π2)2πd)=sin−1 (λ4d)=sin−1 (500 nm×10−9 mm1 mm4(0.120 mm×10−3 m1 mm))=5.97×10−2 degree$

Conclusion:

Therefore, the value of $θ$ is. $5.97×10−2 degree$ .

Answer 8

(a)

Expert Solution

To determine

The phase difference between the two waves fronts arriving at

$P$ when

$θ=0.500°$ .

Answer to Problem 37.18P

The phase difference between the two waves fronts arriving at

$P$ when

$θ=0.500°$ is

$13.2 radian$

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $500 nm$ .

The given diagram is shown below.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 37, Problem 37.18P

Figure 1

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

Substitute $0.120 mm$ for $d$ , $0.500°$ for $θ$ , $500 nm$ for $λ$ , in the above formula as,

$ϕ=2πλdsin θ=2π(500 nm×10−9 m1 nm)(.120 mm×10−3 m1 mm)sin (0.500°)=13.2 radian$

Conclusion:

Therefore, the phase difference between the two waves fronts arriving at $P$ when $θ=0.500°$ is $13.2 radian$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The phase difference between the two waves fronts arriving at

$P$ when

$θ=0.500°$ .

Answer 13

Answer to Problem 37.18P

The phase difference between the two waves fronts arriving at

$P$ when

$θ=0.500°$ is

$13.2 radian$

Answer 14

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $500 nm$ .

The given diagram is shown below.

EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 37, Problem 37.18P

Figure 1

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

Substitute $0.120 mm$ for $d$ , $0.500°$ for $θ$ , $500 nm$ for $λ$ , in the above formula as,

$ϕ=2πλdsin θ=2π(500 nm×10−9 m1 nm)(.120 mm×10−3 m1 mm)sin (0.500°)=13.2 radian$

Conclusion:

Therefore, the phase difference between the two waves fronts arriving at $P$ when $θ=0.500°$ is $13.2 radian$

Answer 15

(b)

Expert Solution

To determine

The phase difference between the two waves fronts arriving at

$P$ when

$y=5.00 mm$ .

Answer to Problem 37.18P

The phase difference between the two waves fronts arriving at

$P$ when

$y=5.00 mm$ is

$6.28 radian$ .

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $6.28 radian$ .

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

From the right angle triangle $sin θ$ is

$sin θ=perpendicular hypotenuse=yL$

Here,

$y$ is the distance from $O$ to $P$ .

$L$ is the distance between slit and screen.

Substitute $yL$ for $sin θ$ in the above formula as,

$ϕ=2πλdsin θ=2πλd(yL)$

Substitute $0.120 mm$ for $d$ , $500 nm$ for $λ$ , $1.20 m$ for $L$ , $5.00 mm$ for $y$ in the above formula as,

$ϕ=2πλdsin θ=2π(500 nm×10−9 m1 nm)(.120 mm×10−3 m1 mm)(5.00 mm×10-3 m1 mm1.20 m)=6.28 radian$

Conclusion:

Therefore, the phase difference between the two waves fronts arriving at $P$ when $y=5.00 mm$ is $6.28 radian$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The phase difference between the two waves fronts arriving at

$P$ when

$y=5.00 mm$ .

Answer 20

Answer to Problem 37.18P

The phase difference between the two waves fronts arriving at

$P$ when

$y=5.00 mm$ is

$6.28 radian$ .

Answer 21

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $6.28 radian$ .

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

From the right angle triangle $sin θ$ is

$sin θ=perpendicular hypotenuse=yL$

Here,

$y$ is the distance from $O$ to $P$ .

$L$ is the distance between slit and screen.

Substitute $yL$ for $sin θ$ in the above formula as,

$ϕ=2πλdsin θ=2πλd(yL)$

Substitute $0.120 mm$ for $d$ , $500 nm$ for $λ$ , $1.20 m$ for $L$ , $5.00 mm$ for $y$ in the above formula as,

$ϕ=2πλdsin θ=2π(500 nm×10−9 m1 nm)(.120 mm×10−3 m1 mm)(5.00 mm×10-3 m1 mm1.20 m)=6.28 radian$

Conclusion:

Therefore, the phase difference between the two waves fronts arriving at $P$ when $y=5.00 mm$ is $6.28 radian$

Answer 22

(c)

Expert Solution

To determine

The value of

$θ$ for which the path difference is

$λ4$ .

Answer to Problem 37.18P

The value of

$θ$ for which the path difference is

$λ4$ is

$1.27×10−2 degree$ .

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $6.28 radian$ , phase difference is $0.33 radian$

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

Rearrange the above formula to find $θ$ is,

$ϕ=2πλdsin θsin θ=λϕ2πdθ=sin−1 (λϕ2πd)$

Substitute $0.120 mm$ for $d$ , $500 nm$ for $λ$ , $0.33 radian$ for $ϕ$ in the above formula as,

$θ=sin−1 (λϕ2πd)=sin−1 (500 nm×10−9 mm1 mm(0.33 radian)2π(0.120 mm×10−3 m1 mm))=1.27×10−2 degree$

Conclusion:

Therefore, the value of $θ$ is $1.27×10−2 degree$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

The value of

$θ$ for which the path difference is

$λ4$ .

Answer 27

Answer to Problem 37.18P

The value of

$θ$ for which the path difference is

$λ4$ is

$1.27×10−2 degree$ .

Answer 28

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $6.28 radian$ , phase difference is $0.33 radian$

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

Rearrange the above formula to find $θ$ is,

$ϕ=2πλdsin θsin θ=λϕ2πdθ=sin−1 (λϕ2πd)$

Substitute $0.120 mm$ for $d$ , $500 nm$ for $λ$ , $0.33 radian$ for $ϕ$ in the above formula as,

$θ=sin−1 (λϕ2πd)=sin−1 (500 nm×10−9 mm1 mm(0.33 radian)2π(0.120 mm×10−3 m1 mm))=1.27×10−2 degree$

Conclusion:

Therefore, the value of $θ$ is $1.27×10−2 degree$ .

Answer 29

(d)

Expert Solution

To determine

The value of

$θ$

Answer to Problem 37.18P

The value of

$θ$ is

$5.97×10−2 degree$ .

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $6.28 radian$ , path difference is $λ4$

The formula to calculate the phase difference is,

$ϕ=2πλdsin θ$ (1)

Here,

$λ$ is the wavelength.

$d$ is the separation between the slits.

$θ$ is the angle between the point $P$ and horizontal.

The path difference is $λ4$ as,

$dsin θ=λ4$

Substitute $λ4$ for $dsin θ$ in the above formula as,

$ϕ=2πλdsin θ=2πλ(λ4)=π2$

Rearrange the equation (1) to find $θ$ as,

$θ=sin−1 (λϕ2πd)$

Substitute $0.120 mm$ for $d$ , $500 nm$ for $λ$ , $π2$ for $ϕ$ in the above formula as,

$θ=sin−1 (λ(π2)2πd)=sin−1 (λ4d)=sin−1 (500 nm×10−9 mm1 mm4(0.120 mm×10−3 m1 mm))=5.97×10−2 degree$

Conclusion:

Therefore, the value of $θ$ is. $5.97×10−2 degree$ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

The value of

$θ$

Answer 34

Answer to Problem 37.18P

The value of

$θ$ is

$5.97×10−2 degree$ .

Answer 35

Explanation of Solution

Given info: The separation between the slits is $0.120 mm$ and the distance between the slit and screen is $1.20 m$ and wavelength of light is $6.28 radian$ , path difference is $λ4$