A ray of light in glass strikes a water-glass interface at an angle of incidence equal to one-half the critical angle for that interface.The index of refraction for water is 1.33,and for the glass it is 1.43.What angle does the refracted ray in the water make with the normal?
A)37°
B)42°
C)47°
D)32°
E)27°

Question

Quizplus · Accepted Answer

The critical angle for the interface can be found using:
sin(critical angle) = n₂/n₁
where n₁ is the refractive index of the first medium (glass) and n₂ is the refractive index of the second medium (water). 
sin(critical angle) = 1.43/1.33
critical angle = sin⁻¹(1.43/1.33) = 51.6°

The angle of incidence for the ray of light is half of the critical angle, so θ = (1/2) × 51.6° = 25.8°.

Using Snell's law, the angle of refraction in water can be found:
n₁sinθ = n₂sinϕ
1.43sin(25.8°) = 1.33sinϕ
sinϕ = (1.43/1.33)sin(25.8°)
sinϕ = 0.833
ϕ = sin⁻¹(0.833) = 37°

Therefore, the refracted ray in water makes an angle of 37° with the normal.

A Ray of Light in Glass Strikes a Water-Glass Interface