Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 23, Problem 74P

(a)

To determine

The ray diagram with all rays emerging from the prism.

(a)

Expert Solution
Check Mark

Answer to Problem 74P

Physics, Chapter 23, Problem 74P , additional homework tip  1

Explanation of Solution

The angle of prism is 45° and refractive index of the glass is 1.50.

Write the law of reflection

  θi=θr                                                                                                                        (I)

Here, θi is the angle of incidence and θr is the angle of reflection.

Physics, Chapter 23, Problem 74P , additional homework tip  2

Write the expression for θi using the small isosceles triangle below the incident ray

  θi+45°=90°                                                                                                           (II)

Find θi using the above equation

  θi=45°

Write the expression for critical angle

  θc=sin1ntni                                                                                                          (III)

Here, θc is the critical angle, nt is the refractive index of the refractive medium and ni is the refractive index of the incident medium.

Substitute 1.00 for nt and 1.50 for nt in (III) to find θc

  θc=sin11.001.50=41.8°

Since θi>θc, there is no transmitted ray because of total internal reflection and the ray travels downward.

(b)

To determine

The ray diagram when the prism is immersed in water.

(b)

Expert Solution
Check Mark

Answer to Problem 74P

The ray diagram when the prism is immersed in water is given in figure 2.

Explanation of Solution

The refractive index of the water is 1.33.

Substitute 1.33 for nt and 1.50 for ni in (III) to find θc

  θc=sin11.001.33=62.5°

Since θi<θc, there is a transmitted ray at an angle given by Snell’s law.

Write the expression for Snell’s law

  n1sinθ1=n2sinθ2                                                                                             (IV)

Here, θ1 is the incident angle, θ2 is the angle of refraction, n1 is the refractive index of the refractive medium and n2 is the refractive index of the incident medium.

Rearrange for θ2

  θ2=sin1(n1n2sinθ1)                                                                                            (V)

Substitute 1.50 for n1, 1.33 for n2 and 45° for θ1 in (V) to find θ2

  θ2=sin1(1.501.33sin45°)=53°

Thus, the angle of refraction is 53°.

Physics, Chapter 23, Problem 74P , additional homework tip  3

(c)

To determine

The ray diagram when the prism is immersed in sugar solution.

(c)

Expert Solution
Check Mark

Answer to Problem 74P

The ray diagram when the prism is immersed in sugar solution is given in figure 3.

Explanation of Solution

The refractive index of the sugar solution is 1.50 which is same as the refractive index of the prism. Hence no refraction occurs and the ray emerges without any change in direction.

The ray diagram for prism immersed in sugar solution is given below

Physics, Chapter 23, Problem 74P , additional homework tip  4

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Chapter 23 Solutions

Physics

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Laws of Refraction of Light | Don't Memorise; Author: Don't Memorise;https://www.youtube.com/watch?v=4l2thi5_84o;License: Standard YouTube License, CC-BY