Question

A triangular glass prism with apex angle of 60.0° has an index of refraction n=1.50. What is the smallest angle of incidence for which a light ray can emerge from the other side if the prism is surrounded by (i) air (n=1.00), (ii) water (n=1.33)?

Answer 1

(i) prism is surroumded by air (n =1)

Step:1 Finding the critical angle

Sin_c = 1/1.5

_c = sin^-1(1/1.5) = 41.81°

Step:2 Solve using geometry

90-41.81 = 48.19

180-60-48.19 = 71.81

₂ = 90 -71.81 = 18.19°

Step:3 Using snell's law

n₁sin₁ = n₂sin₂ , where ₁is the smallest angle of incidence

(1)sin₁ = (1.5)sin(18.19°)

Sin₁ = 0.468

i.e. ₁ = 27.92°

(ii) The prism is surrounded by water (n= 1.33)

step:1 Finding the critical angle

sin_c = 1.33/1.5

_c = 62.46°

step:2 solve using geometry

90 - 62.46 = 27.54

180 - 60 - 27.54 = 92.46

92.46 - 90 = 2.46

step:3 using snell law:

(1.33)sin₁ = 1.5 sin(2.46)

₁ = 2.77