Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 23, Problem 85P

(a)

To determine

The refractive index of the unknown substance.

(a)

Expert Solution
Check Mark

Answer to Problem 85P

The refractive index of the unknown substance is 1.42_.

Explanation of Solution

Given that the angle of reflection is 25.0°, angle of refraction is 37.0°.

According to the laws of reflection the angle of incidence equals the angle of reflection.

Write the expression of Snell’s law of refraction.

    n1sinθ1=n2sinθ2                                                                                                     (I)

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

According to the law of reflection, the angle of incidence is equal to the angle of reflection. Thus, in the given case, the angle of incidence is 25.0°. In the given system, the first medium is the unknown substance and the second medium is air (n2=1.000).

Solve equation (I) for n1.

    n1=n2sinθ2sinθ1                                                                                                           (II)

Conclusion:

Substitute 1.000 n2, 37.0° for θ2 and 25.0° for θ1 in the equation (IV) to find n1.

    n1=1.000sin37.0°sin25°=1.42

Therefore, the refractive index of the unknown substance is 1.42_.

(b)

To determine

The speed of light in the unknown substance.

(b)

Expert Solution
Check Mark

Answer to Problem 85P

The speed of the light in the unknown substance is 2.11×108m/s_.

Explanation of Solution

It is obtained that the refractive index of the unknown substance is 1.42.

Write the expression for the speed of light in a medium

    v=cn                                                                                                                      (III)

Here, v is the speed of the light in the medium, c is the speed of light in vacuum, n is the refractive index of the medium.

Conclusion:

Substitute 3.00×108m/s for c, 1.42 for n in equation (III) to find the speed of light in the unknown substance.

    v=3.00×108m/s1.42=2.11×108m/s

Therefore, the speed of the light in the unknown substance is 2.11×108m/s_.

(c)

To determine

The minimum angle of incidence at which the light would undergo total internal reflection.

(c)

Expert Solution
Check Mark

Answer to Problem 85P

The minimum angle of incidence at which the light would undergo total internal reflection is 44.6°_.

Explanation of Solution

When a light passes from a high refractive index medium to a lower refractive index medium, and if the angle of incidence is greater than the critical angle, the total internal reflection occurs.

Write the expression for the critical angle.

    θc=sin1n2n1                                                                                                           (IV)

Here, θc is the critical angle.

Use equation (II) in equation (IV).

    θc=sin1n2(n2sinθ2sinθ1)=sin1sinθ1sinθ2                                                                                              (V)

Conclusion:

Substitute 25.0° for θ1 and 37.0° for θ2 in equation (VII) to find θc.

    θc=sin1(sin25.0°sin37.0°)=44.6°

Therefore, the minimum angle of incidence at which the light would undergo total internal reflection is 44.6°_.

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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