A plate of glass with parallel faces having a refractive index of 1.57 is resting on the surface of water in a tank. A ray of light coming from above in air makes an angle of incidence 37.5 ∘∘ with the normal to the top surface of the glass. What angle does the ray refracted into the water make with the normal to the surface? Use 1.33 for the index of refraction of water.

The angle that the ray refracted into the water makes with the normal to the surface is; r = 27.24°

We are given;

Refractive index of glass; η_g = 1.57

Refractive index of water; η_w = 1.33

Angle of incidence; i = 37.5°

First of all, in air;

Refractive index of air is; η_a = 1

We will use snell's law to find the angle of refraction. Thus;

η_g/η_a = sin i/sin r

∴ 1.57/1 = (sin 37.5)/sin r

sin r = 0.3877

r = sin⁻¹0.3877

r = 22.81°

The angle of 22.81° will now be the incident angle for the water medium. Thus;

η_w/η_g = sin i/sin r

1.33/1.57 = sin 22.81/sin r

sin r = 0.3877/0.8471

sin r = 0.4577

r = sin⁻¹0.4577

r = 27.24°

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Step-by-step explanation:

We are given that

Angle of incidence,i=37.5 degree

We have to find the angle of refraction into the water made by ray wit the normal to the surface.

Refractive index of water,

By Snell's law

Again for glass water surface

Using Snell's law

Angle of incidence,i'=22.83 degree

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