Answers to Self-Assessment Problems - Geometrical and Visual Optics - page 267

Answers to Self-Assessment
Problems

CHAPTER 1
Basic Terms and Concepts
1. n sin θ = n ′ sin θ ′

(1.33) sin θ = (1.00) sin 45°

θ = 32.1°

2. n sin θ = n ′ sin θ ′

(1.00) sin 25° = (1.33) sin θ ′

θ ′ = 18.53°

If the ray was not deviated:

d
200.0 cm

tan 25 ° =

d = 93.26 cm

But because the ray is refracted:

d
200.0 cm

tan 18.53 ° =

d = 67.04 cm

Therefore, the deviation of the ray is

93.26 − 67.04 cm = 26.22 cm

3. At the first interface:

n sin θ = n ′ sin θ ′

(1.00) sin30° = 1.52 sin θ ′

θ ′ = 19.2°

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