Light is incident on a prism as shown in Figure P38.31. The prism, an equilateral triangle, is made of plastic with an index of refraction of 1.46 for red light and 1.49 for blue light. Assume the apex angle of the prism is 60.00°. a. Sketch the approximate paths of the rays for red and blue light as they travel through and then exit the prism. b. Determine the measure of dispersion, the angle between the red and blue rays that exit the prism. Figure P38.31

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Answer 1

Textbook Question

Chapter 38, Problem 31PQ

Light is incident on a prism as shown in Figure P38.31. The prism, an equilateral triangle, is made of plastic with an index of refraction of 1.46 for red light and 1.49 for blue light. Assume the apex angle of the prism is 60.00°.

a. Sketch the approximate paths of the rays for red and blue light as they travel through and then exit the prism.
b. Determine the measure of dispersion, the angle between the red and blue rays that exit the prism.

Chapter 38, Problem 31PQ, Light is incident on a prism as shown in Figure P38.31. The prism, an equilateral triangle, is made

Figure P38.31

(a)

Expert Solution

To determine

The sketch of the appropriate path of the red and blue light rays through the prism.

Answer to Problem 31PQ

The sketch of the appropriate path of the red and blue light rays through the prism is as shown below.

Physics for Scientists and Engineers: Foundations and Connections, Advance Edition, Volume 2, Chapter 38, Problem 31PQ , additional homework tip 1

Explanation of Solution

The wavelength of the violet light is $410 nm$ and the wavelength of the red light is $680 nm$ .

The wavelength of light is inversely proportion to the refractive index of the material through which it is passing.

The wavelength of the violet light is lesser than the wavelength of the red light. Hence, the refractive index of the violet light is more as compared to the refractive index for the red light.

Therefore, violet light refracts more as compared to the red light as shown in the figure below.

Physics for Scientists and Engineers: Foundations and Connections, Advance Edition, Volume 2, Chapter 38, Problem 31PQ , additional homework tip 2

Figure-(1)

(b)

Expert Solution

To determine

The dispersion angle between the red and blue rays when exit the prism.

Answer to Problem 31PQ

The dispersion angle between the red and blue rays when exit the prism is $2.64°$ .

Explanation of Solution

Write the expression for Snell’s law.

$n1sinθi=n2sinθt$

Here, incidence angle is $θi$ , refractive index of the first medium is $n1$ , refraction angle is $θt$ and the refractive index of the second medium is $n2$ .

Further solve the above equation for $θt$ .

$sinθt=n1n2sinθiθt=sin−1(n1n2sinθi)$ (I)

The prism diagram on which light is incident at an angle $45°$ is as shown below.

Physics for Scientists and Engineers: Foundations and Connections, Advance Edition, Volume 2, Chapter 38, Problem 31PQ , additional homework tip 3

Figure-(1)

Write the expression for dispersion angle between the red and blue rays when exit the prism.

$ϕ=(θt')b−(θt')r$ (II)

Here, dispersion angle between the red and blue rays when exit the prism is $ϕ$ , refracted angle at the exit of the prism for blue light is $(θt')b$ and refracted angle at the exit of the prism for red light is $(θt')r$ .

Conclusion:

Case (i): For blue light.

Substitute $1$ for $n1$ , $1.49$ for $n2$ and $45.0°$ for $θi$ in equation (I) to find $(θt)b$

$(θt)b=sin−1(11.49sin45°)=sin−1(0.474)=28.33°$

Solve for $α$ .

$α=90°−(θt)b=90°−28.33°=61.67°$

Solve for $β$ .

$β=180°−60°−α=180°−60°−61.67°=58.33°$

Solve for $θi'$ .

$θi'=90°−β=90°−58.33°=31.67°$

Substitute $31.67°$ for $θi'$ , $1.49$ for $n1$ and $1$ for $n2$ in equation (I) to find $(θj')b$ .

$(θt')b=sin−1(1.491sin31.67°)=sin−1(0.782)=51.47°$

Case (ii): For red light.

Substitute $1$ for $n1$ , $1.46$ for $n2$ and $45.0°$ for $θi$ in equation (I) to find $(θt)b$

$(θt)r=sin−1(11.46sin45°)=sin−1(0.484)=28.96°$

Solve for $α$ .

$α=90°−(θt)b=90°−28.96°=61.04°$

Solve for $β$ .

$β=180°−60°−α=180°−60°−61.04°=58.96°$

Solve for $θi'$ .

$θi'=90°−β=90°−58.69°=31.04°$

Substitute $31.04°$ for $θi'$ , $1.49$ for $n1$ and $1$ for $n2$ in equation (I) to find $(θj')r$ .

$(θj')r=sin−1(1.461sin31.04°)=sin−1(0.752)=48.83°$

Substitute $51.47°$ for $(θt')b$ and $48.83°$ for $(θj')r$ in equation (II) to find $ϕ$ .

$ϕ=51.47°−48.83°=2.64°$

Therefore, the dispersion angle between the red and blue rays when exit the prism is $2.64°$ .

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Answer 2

Textbook Question

Chapter 38, Problem 31PQ

Light is incident on a prism as shown in Figure P38.31. The prism, an equilateral triangle, is made of plastic with an index of refraction of 1.46 for red light and 1.49 for blue light. Assume the apex angle of the prism is 60.00°.

a. Sketch the approximate paths of the rays for red and blue light as they travel through and then exit the prism.
b. Determine the measure of dispersion, the angle between the red and blue rays that exit the prism.

Chapter 38, Problem 31PQ, Light is incident on a prism as shown in Figure P38.31. The prism, an equilateral triangle, is made

Figure P38.31

(a)

Expert Solution

To determine

The sketch of the appropriate path of the red and blue light rays through the prism.

Answer to Problem 31PQ

The sketch of the appropriate path of the red and blue light rays through the prism is as shown below.

Physics for Scientists and Engineers: Foundations and Connections, Advance Edition, Volume 2, Chapter 38, Problem 31PQ , additional homework tip 1

Explanation of Solution

The wavelength of the violet light is $410 nm$ and the wavelength of the red light is $680 nm$ .

The wavelength of light is inversely proportion to the refractive index of the material through which it is passing.

The wavelength of the violet light is lesser than the wavelength of the red light. Hence, the refractive index of the violet light is more as compared to the refractive index for the red light.

Therefore, violet light refracts more as compared to the red light as shown in the figure below.

Physics for Scientists and Engineers: Foundations and Connections, Advance Edition, Volume 2, Chapter 38, Problem 31PQ , additional homework tip 2

Figure-(1)

(b)

Expert Solution

To determine

The dispersion angle between the red and blue rays when exit the prism.

Answer to Problem 31PQ

The dispersion angle between the red and blue rays when exit the prism is $2.64°$ .

Explanation of Solution

Write the expression for Snell’s law.

$n1sinθi=n2sinθt$

Here, incidence angle is $θi$ , refractive index of the first medium is $n1$ , refraction angle is $θt$ and the refractive index of the second medium is $n2$ .

Further solve the above equation for $θt$ .

$sinθt=n1n2sinθiθt=sin−1(n1n2sinθi)$ (I)

The prism diagram on which light is incident at an angle $45°$ is as shown below.

Physics for Scientists and Engineers: Foundations and Connections, Advance Edition, Volume 2, Chapter 38, Problem 31PQ , additional homework tip 3

Figure-(1)

Write the expression for dispersion angle between the red and blue rays when exit the prism.

$ϕ=(θt')b−(θt')r$ (II)

Here, dispersion angle between the red and blue rays when exit the prism is $ϕ$ , refracted angle at the exit of the prism for blue light is $(θt')b$ and refracted angle at the exit of the prism for red light is $(θt')r$ .

Conclusion:

Case (i): For blue light.

Substitute $1$ for $n1$ , $1.49$ for $n2$ and $45.0°$ for $θi$ in equation (I) to find $(θt)b$

$(θt)b=sin−1(11.49sin45°)=sin−1(0.474)=28.33°$

Solve for $α$ .

$α=90°−(θt)b=90°−28.33°=61.67°$

Solve for $β$ .

$β=180°−60°−α=180°−60°−61.67°=58.33°$

Solve for $θi'$ .

$θi'=90°−β=90°−58.33°=31.67°$

Substitute $31.67°$ for $θi'$ , $1.49$ for $n1$ and $1$ for $n2$ in equation (I) to find $(θj')b$ .

$(θt')b=sin−1(1.491sin31.67°)=sin−1(0.782)=51.47°$

Case (ii): For red light.

Substitute $1$ for $n1$ , $1.46$ for $n2$ and $45.0°$ for $θi$ in equation (I) to find $(θt)b$

$(θt)r=sin−1(11.46sin45°)=sin−1(0.484)=28.96°$

Solve for $α$ .

$α=90°−(θt)b=90°−28.96°=61.04°$

Solve for $β$ .

$β=180°−60°−α=180°−60°−61.04°=58.96°$

Solve for $θi'$ .

$θi'=90°−β=90°−58.69°=31.04°$

Substitute $31.04°$ for $θi'$ , $1.49$ for $n1$ and $1$ for $n2$ in equation (I) to find $(θj')r$ .

$(θj')r=sin−1(1.461sin31.04°)=sin−1(0.752)=48.83°$

Substitute $51.47°$ for $(θt')b$ and $48.83°$ for $(θj')r$ in equation (II) to find $ϕ$ .

$ϕ=51.47°−48.83°=2.64°$

Therefore, the dispersion angle between the red and blue rays when exit the prism is $2.64°$ .

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Answer 3

Textbook Question

Chapter 38, Problem 31PQ

Light is incident on a prism as shown in Figure P38.31. The prism, an equilateral triangle, is made of plastic with an index of refraction of 1.46 for red light and 1.49 for blue light. Assume the apex angle of the prism is 60.00°.

a. Sketch the approximate paths of the rays for red and blue light as they travel through and then exit the prism.
b. Determine the measure of dispersion, the angle between the red and blue rays that exit the prism.

Chapter 38, Problem 31PQ, Light is incident on a prism as shown in Figure P38.31. The prism, an equilateral triangle, is made

Figure P38.31

(a)

Expert Solution

To determine

The sketch of the appropriate path of the red and blue light rays through the prism.

Answer to Problem 31PQ

The sketch of the appropriate path of the red and blue light rays through the prism is as shown below.

Physics for Scientists and Engineers: Foundations and Connections, Advance Edition, Volume 2, Chapter 38, Problem 31PQ , additional homework tip 1

Explanation of Solution

The wavelength of the violet light is $410 nm$ and the wavelength of the red light is $680 nm$ .

The wavelength of light is inversely proportion to the refractive index of the material through which it is passing.

The wavelength of the violet light is lesser than the wavelength of the red light. Hence, the refractive index of the violet light is more as compared to the refractive index for the red light.

Therefore, violet light refracts more as compared to the red light as shown in the figure below.

Physics for Scientists and Engineers: Foundations and Connections, Advance Edition, Volume 2, Chapter 38, Problem 31PQ , additional homework tip 2

Figure-(1)

(b)

Expert Solution

To determine

The dispersion angle between the red and blue rays when exit the prism.

Answer to Problem 31PQ

The dispersion angle between the red and blue rays when exit the prism is $2.64°$ .

Explanation of Solution

Write the expression for Snell’s law.

$n1sinθi=n2sinθt$

Here, incidence angle is $θi$ , refractive index of the first medium is $n1$ , refraction angle is $θt$ and the refractive index of the second medium is $n2$ .

Further solve the above equation for $θt$ .

$sinθt=n1n2sinθiθt=sin−1(n1n2sinθi)$ (I)

The prism diagram on which light is incident at an angle $45°$ is as shown below.

Physics for Scientists and Engineers: Foundations and Connections, Advance Edition, Volume 2, Chapter 38, Problem 31PQ , additional homework tip 3

Figure-(1)

Write the expression for dispersion angle between the red and blue rays when exit the prism.

$ϕ=(θt')b−(θt')r$ (II)

Here, dispersion angle between the red and blue rays when exit the prism is $ϕ$ , refracted angle at the exit of the prism for blue light is $(θt')b$ and refracted angle at the exit of the prism for red light is $(θt')r$ .

Conclusion:

Case (i): For blue light.

Substitute $1$ for $n1$ , $1.49$ for $n2$ and $45.0°$ for $θi$ in equation (I) to find $(θt)b$

$(θt)b=sin−1(11.49sin45°)=sin−1(0.474)=28.33°$

Solve for $α$ .

$α=90°−(θt)b=90°−28.33°=61.67°$

Solve for $β$ .

$β=180°−60°−α=180°−60°−61.67°=58.33°$

Solve for $θi'$ .

$θi'=90°−β=90°−58.33°=31.67°$

Substitute $31.67°$ for $θi'$ , $1.49$ for $n1$ and $1$ for $n2$ in equation (I) to find $(θj')b$ .

$(θt')b=sin−1(1.491sin31.67°)=sin−1(0.782)=51.47°$

Case (ii): For red light.

Substitute $1$ for $n1$ , $1.46$ for $n2$ and $45.0°$ for $θi$ in equation (I) to find $(θt)b$

$(θt)r=sin−1(11.46sin45°)=sin−1(0.484)=28.96°$

Solve for $α$ .

$α=90°−(θt)b=90°−28.96°=61.04°$

Solve for $β$ .

$β=180°−60°−α=180°−60°−61.04°=58.96°$

Solve for $θi'$ .

$θi'=90°−β=90°−58.69°=31.04°$

Substitute $31.04°$ for $θi'$ , $1.49$ for $n1$ and $1$ for $n2$ in equation (I) to find $(θj')r$ .

$(θj')r=sin−1(1.461sin31.04°)=sin−1(0.752)=48.83°$

Substitute $51.47°$ for $(θt')b$ and $48.83°$ for $(θj')r$ in equation (II) to find $ϕ$ .

$ϕ=51.47°−48.83°=2.64°$

Therefore, the dispersion angle between the red and blue rays when exit the prism is $2.64°$ .

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Answer 4

Textbook Question

Chapter 38, Problem 31PQ

Light is incident on a prism as shown in Figure P38.31. The prism, an equilateral triangle, is made of plastic with an index of refraction of 1.46 for red light and 1.49 for blue light. Assume the apex angle of the prism is 60.00°.

a. Sketch the approximate paths of the rays for red and blue light as they travel through and then exit the prism.
b. Determine the measure of dispersion, the angle between the red and blue rays that exit the prism.

Chapter 38, Problem 31PQ, Light is incident on a prism as shown in Figure P38.31. The prism, an equilateral triangle, is made

Figure P38.31

(a)

Expert Solution

To determine

The sketch of the appropriate path of the red and blue light rays through the prism.

Answer to Problem 31PQ

The sketch of the appropriate path of the red and blue light rays through the prism is as shown below.

Physics for Scientists and Engineers: Foundations and Connections, Advance Edition, Volume 2, Chapter 38, Problem 31PQ , additional homework tip 1

Explanation of Solution

The wavelength of the violet light is $410 nm$ and the wavelength of the red light is $680 nm$ .

The wavelength of light is inversely proportion to the refractive index of the material through which it is passing.

The wavelength of the violet light is lesser than the wavelength of the red light. Hence, the refractive index of the violet light is more as compared to the refractive index for the red light.

Therefore, violet light refracts more as compared to the red light as shown in the figure below.

Physics for Scientists and Engineers: Foundations and Connections, Advance Edition, Volume 2, Chapter 38, Problem 31PQ , additional homework tip 2

Figure-(1)

(b)

Expert Solution

To determine

The dispersion angle between the red and blue rays when exit the prism.

Answer to Problem 31PQ

The dispersion angle between the red and blue rays when exit the prism is $2.64°$ .

Explanation of Solution

Write the expression for Snell’s law.

$n1sinθi=n2sinθt$

Here, incidence angle is $θi$ , refractive index of the first medium is $n1$ , refraction angle is $θt$ and the refractive index of the second medium is $n2$ .

Further solve the above equation for $θt$ .

$sinθt=n1n2sinθiθt=sin−1(n1n2sinθi)$ (I)

The prism diagram on which light is incident at an angle $45°$ is as shown below.

Physics for Scientists and Engineers: Foundations and Connections, Advance Edition, Volume 2, Chapter 38, Problem 31PQ , additional homework tip 3

Figure-(1)

Write the expression for dispersion angle between the red and blue rays when exit the prism.

$ϕ=(θt')b−(θt')r$ (II)

Here, dispersion angle between the red and blue rays when exit the prism is $ϕ$ , refracted angle at the exit of the prism for blue light is $(θt')b$ and refracted angle at the exit of the prism for red light is $(θt')r$ .

Conclusion:

Case (i): For blue light.

Substitute $1$ for $n1$ , $1.49$ for $n2$ and $45.0°$ for $θi$ in equation (I) to find $(θt)b$

$(θt)b=sin−1(11.49sin45°)=sin−1(0.474)=28.33°$

Solve for $α$ .

$α=90°−(θt)b=90°−28.33°=61.67°$

Solve for $β$ .

$β=180°−60°−α=180°−60°−61.67°=58.33°$

Solve for $θi'$ .

$θi'=90°−β=90°−58.33°=31.67°$

Substitute $31.67°$ for $θi'$ , $1.49$ for $n1$ and $1$ for $n2$ in equation (I) to find $(θj')b$ .

$(θt')b=sin−1(1.491sin31.67°)=sin−1(0.782)=51.47°$

Case (ii): For red light.

Substitute $1$ for $n1$ , $1.46$ for $n2$ and $45.0°$ for $θi$ in equation (I) to find $(θt)b$

$(θt)r=sin−1(11.46sin45°)=sin−1(0.484)=28.96°$

Solve for $α$ .

$α=90°−(θt)b=90°−28.96°=61.04°$

Solve for $β$ .

$β=180°−60°−α=180°−60°−61.04°=58.96°$

Solve for $θi'$ .

$θi'=90°−β=90°−58.69°=31.04°$

Substitute $31.04°$ for $θi'$ , $1.49$ for $n1$ and $1$ for $n2$ in equation (I) to find $(θj')r$ .

$(θj')r=sin−1(1.461sin31.04°)=sin−1(0.752)=48.83°$

Substitute $51.47°$ for $(θt')b$ and $48.83°$ for $(θj')r$ in equation (II) to find $ϕ$ .

$ϕ=51.47°−48.83°=2.64°$

Therefore, the dispersion angle between the red and blue rays when exit the prism is $2.64°$ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The sketch of the appropriate path of the red and blue light rays through the prism.

Answer to Problem 31PQ

The sketch of the appropriate path of the red and blue light rays through the prism is as shown below.

Physics for Scientists and Engineers: Foundations and Connections, Advance Edition, Volume 2, Chapter 38, Problem 31PQ , additional homework tip 1

Explanation of Solution

The wavelength of the violet light is $410 nm$ and the wavelength of the red light is $680 nm$ .

The wavelength of light is inversely proportion to the refractive index of the material through which it is passing.

The wavelength of the violet light is lesser than the wavelength of the red light. Hence, the refractive index of the violet light is more as compared to the refractive index for the red light.

Therefore, violet light refracts more as compared to the red light as shown in the figure below.

Physics for Scientists and Engineers: Foundations and Connections, Advance Edition, Volume 2, Chapter 38, Problem 31PQ , additional homework tip 2

Figure-(1)

(b)

Expert Solution

To determine

The dispersion angle between the red and blue rays when exit the prism.

Answer to Problem 31PQ

The dispersion angle between the red and blue rays when exit the prism is $2.64°$ .

Explanation of Solution

Write the expression for Snell’s law.

$n1sinθi=n2sinθt$

Here, incidence angle is $θi$ , refractive index of the first medium is $n1$ , refraction angle is $θt$ and the refractive index of the second medium is $n2$ .

Further solve the above equation for $θt$ .

$sinθt=n1n2sinθiθt=sin−1(n1n2sinθi)$ (I)

The prism diagram on which light is incident at an angle $45°$ is as shown below.

Physics for Scientists and Engineers: Foundations and Connections, Advance Edition, Volume 2, Chapter 38, Problem 31PQ , additional homework tip 3

Figure-(1)

Write the expression for dispersion angle between the red and blue rays when exit the prism.

$ϕ=(θt')b−(θt')r$ (II)

Here, dispersion angle between the red and blue rays when exit the prism is $ϕ$ , refracted angle at the exit of the prism for blue light is $(θt')b$ and refracted angle at the exit of the prism for red light is $(θt')r$ .

Conclusion:

Case (i): For blue light.

Substitute $1$ for $n1$ , $1.49$ for $n2$ and $45.0°$ for $θi$ in equation (I) to find $(θt)b$

$(θt)b=sin−1(11.49sin45°)=sin−1(0.474)=28.33°$

Solve for $α$ .

$α=90°−(θt)b=90°−28.33°=61.67°$

Solve for $β$ .

$β=180°−60°−α=180°−60°−61.67°=58.33°$

Solve for $θi'$ .

$θi'=90°−β=90°−58.33°=31.67°$

Substitute $31.67°$ for $θi'$ , $1.49$ for $n1$ and $1$ for $n2$ in equation (I) to find $(θj')b$ .

$(θt')b=sin−1(1.491sin31.67°)=sin−1(0.782)=51.47°$

Case (ii): For red light.

Substitute $1$ for $n1$ , $1.46$ for $n2$ and $45.0°$ for $θi$ in equation (I) to find $(θt)b$

$(θt)r=sin−1(11.46sin45°)=sin−1(0.484)=28.96°$

Solve for $α$ .

$α=90°−(θt)b=90°−28.96°=61.04°$

Solve for $β$ .

$β=180°−60°−α=180°−60°−61.04°=58.96°$

Solve for $θi'$ .

$θi'=90°−β=90°−58.69°=31.04°$

Substitute $31.04°$ for $θi'$ , $1.49$ for $n1$ and $1$ for $n2$ in equation (I) to find $(θj')r$ .

$(θj')r=sin−1(1.461sin31.04°)=sin−1(0.752)=48.83°$

Substitute $51.47°$ for $(θt')b$ and $48.83°$ for $(θj')r$ in equation (II) to find $ϕ$ .

$ϕ=51.47°−48.83°=2.64°$

Therefore, the dispersion angle between the red and blue rays when exit the prism is $2.64°$ .

Answer 8

(a)

Expert Solution

To determine

The sketch of the appropriate path of the red and blue light rays through the prism.

Answer to Problem 31PQ

The sketch of the appropriate path of the red and blue light rays through the prism is as shown below.

Physics for Scientists and Engineers: Foundations and Connections, Advance Edition, Volume 2, Chapter 38, Problem 31PQ , additional homework tip 1

Explanation of Solution

The wavelength of the violet light is $410 nm$ and the wavelength of the red light is $680 nm$ .

The wavelength of light is inversely proportion to the refractive index of the material through which it is passing.

The wavelength of the violet light is lesser than the wavelength of the red light. Hence, the refractive index of the violet light is more as compared to the refractive index for the red light.

Therefore, violet light refracts more as compared to the red light as shown in the figure below.

Physics for Scientists and Engineers: Foundations and Connections, Advance Edition, Volume 2, Chapter 38, Problem 31PQ , additional homework tip 2

Figure-(1)

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The sketch of the appropriate path of the red and blue light rays through the prism.

Answer 13

Answer to Problem 31PQ

The sketch of the appropriate path of the red and blue light rays through the prism is as shown below.

Physics for Scientists and Engineers: Foundations and Connections, Advance Edition, Volume 2, Chapter 38, Problem 31PQ , additional homework tip 1

Answer 14

Explanation of Solution

The wavelength of the violet light is $410 nm$ and the wavelength of the red light is $680 nm$ .

The wavelength of light is inversely proportion to the refractive index of the material through which it is passing.

The wavelength of the violet light is lesser than the wavelength of the red light. Hence, the refractive index of the violet light is more as compared to the refractive index for the red light.

Therefore, violet light refracts more as compared to the red light as shown in the figure below.

Physics for Scientists and Engineers: Foundations and Connections, Advance Edition, Volume 2, Chapter 38, Problem 31PQ , additional homework tip 2

Figure-(1)

Answer 15

(b)

Expert Solution

To determine

The dispersion angle between the red and blue rays when exit the prism.

Answer to Problem 31PQ

The dispersion angle between the red and blue rays when exit the prism is $2.64°$ .

Explanation of Solution

Write the expression for Snell’s law.

$n1sinθi=n2sinθt$

Here, incidence angle is $θi$ , refractive index of the first medium is $n1$ , refraction angle is $θt$ and the refractive index of the second medium is $n2$ .

Further solve the above equation for $θt$ .

$sinθt=n1n2sinθiθt=sin−1(n1n2sinθi)$ (I)

The prism diagram on which light is incident at an angle $45°$ is as shown below.

Physics for Scientists and Engineers: Foundations and Connections, Advance Edition, Volume 2, Chapter 38, Problem 31PQ , additional homework tip 3

Figure-(1)

Write the expression for dispersion angle between the red and blue rays when exit the prism.

$ϕ=(θt')b−(θt')r$ (II)

Here, dispersion angle between the red and blue rays when exit the prism is $ϕ$ , refracted angle at the exit of the prism for blue light is $(θt')b$ and refracted angle at the exit of the prism for red light is $(θt')r$ .

Conclusion:

Case (i): For blue light.

Substitute $1$ for $n1$ , $1.49$ for $n2$ and $45.0°$ for $θi$ in equation (I) to find $(θt)b$

$(θt)b=sin−1(11.49sin45°)=sin−1(0.474)=28.33°$

Solve for $α$ .

$α=90°−(θt)b=90°−28.33°=61.67°$

Solve for $β$ .

$β=180°−60°−α=180°−60°−61.67°=58.33°$

Solve for $θi'$ .

$θi'=90°−β=90°−58.33°=31.67°$

Substitute $31.67°$ for $θi'$ , $1.49$ for $n1$ and $1$ for $n2$ in equation (I) to find $(θj')b$ .

$(θt')b=sin−1(1.491sin31.67°)=sin−1(0.782)=51.47°$

Case (ii): For red light.

Substitute $1$ for $n1$ , $1.46$ for $n2$ and $45.0°$ for $θi$ in equation (I) to find $(θt)b$

$(θt)r=sin−1(11.46sin45°)=sin−1(0.484)=28.96°$

Solve for $α$ .

$α=90°−(θt)b=90°−28.96°=61.04°$

Solve for $β$ .

$β=180°−60°−α=180°−60°−61.04°=58.96°$

Solve for $θi'$ .

$θi'=90°−β=90°−58.69°=31.04°$

Substitute $31.04°$ for $θi'$ , $1.49$ for $n1$ and $1$ for $n2$ in equation (I) to find $(θj')r$ .

$(θj')r=sin−1(1.461sin31.04°)=sin−1(0.752)=48.83°$

Substitute $51.47°$ for $(θt')b$ and $48.83°$ for $(θj')r$ in equation (II) to find $ϕ$ .

$ϕ=51.47°−48.83°=2.64°$

Therefore, the dispersion angle between the red and blue rays when exit the prism is $2.64°$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The dispersion angle between the red and blue rays when exit the prism.

Answer 20

Answer to Problem 31PQ

The dispersion angle between the red and blue rays when exit the prism is $2.64°$ .

Answer 21

Explanation of Solution

Write the expression for Snell’s law.

$n1sinθi=n2sinθt$

Here, incidence angle is $θi$ , refractive index of the first medium is $n1$ , refraction angle is $θt$ and the refractive index of the second medium is $n2$ .

Further solve the above equation for $θt$ .

$sinθt=n1n2sinθiθt=sin−1(n1n2sinθi)$ (I)

The prism diagram on which light is incident at an angle $45°$ is as shown below.

Physics for Scientists and Engineers: Foundations and Connections, Advance Edition, Volume 2, Chapter 38, Problem 31PQ , additional homework tip 3

Figure-(1)

Write the expression for dispersion angle between the red and blue rays when exit the prism.

$ϕ=(θt')b−(θt')r$ (II)

Here, dispersion angle between the red and blue rays when exit the prism is $ϕ$ , refracted angle at the exit of the prism for blue light is $(θt')b$ and refracted angle at the exit of the prism for red light is $(θt')r$ .