Explanation If looking straight down from air inti a clear, the bottom of the liquid appears at an apparent depth, it is smaller than the actual length. Consider the situation that a transparent material with refractive index n 2 and a curved surface of radius R . Outside the material is a medium with refractive index n 1 . The rays from a point object are refracted to a common point. Consider the Figure 1 below, it shows a single ray from the object. Apply Snell’s law at the surface. n 1 sin θ 1 = n 2 sin θ 2 (I) Here, θ 1 is the angle of incidence, and θ 2 is the angle of refraction. Simplify the equation (II). n 1 θ 1 = n 2 θ 2 (II) Since the angle considered as small. From the triangle B C I in the Figure, θ 2 can be written as, θ 2 = β − γ (III) From the triangle O B C in the Figure, θ 1 can be written as, θ 1 = α + β (IV) Use equation (III) and (IV) in (II). n 1 ( α + β ) = n 2 ( β − γ ) (V) From the triangle O B C in the Figure, write the expression for α . α ≃ tan α ≃ B A s (VI) Here, s is the objects’ distance from the refracting surface. From the triangle O B C in the Figure, write the expression for β . β ≃ B A R (VII) From the triangle B A I in the Figure, write the expression for γ . γ ≃ B A s ′ (VIII) Use equation (VI), (VII), and (VIII) in (V)

4th Edition

ISBN: 9780134988566

Author: Wolfson, Richard

Publisher: Pearson Education,

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Chapter 31, Problem 22E

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The expression for apparent depth in terms of actual depth and refractive index.

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Chapter 31, Problem 21E

Chapter 31, Problem 23E

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