Question

Figures (a) and (b) show the refraction of a ray in air incident at 60° with the normal to a glass-air and water-air interface, respectively. Predict the angle of refraction in the glass when the angle of incidence in water is 45° with the normal to a water-glass interface [Figure (c)].

(a)	(b)	(c)

Answer 1

As per the given figure, for the glass-air interface:

Angle of incidence, i = 60°

Angle of refraction, r = 35°

The relative refractive index of glass with respect to air is given by Snell’s law as:

`""^"a"μ_"g" = (sin "i")/(sin "r")`

= `(sin 60°)/(sin 35°)`

= `(0.8660)/(0.5736)`

= 1.51 ......(1)

As per the given figure, for the air-water interface:

Angle of incidence, i = 60°

Angle of refraction, r = 47°

The relative refractive index of water with respect to air is given by Snell’s law as:

`""^"a"μ_"w" = (sin "i")/(sin "r")`

= `(sin 60°)/(sin 47°)`

= `(0.8660)/(0.7314)`

= 1.184 ....(2)

Using (1) and (2), the relative refractive index of glass with respect to water can be obtained as:

`""^"w"μ_"g" = (""^"a"μ_"g")/(""^"a"μ_"w")`

= `1.51/1.184`

= 1.275

The following figure shows the situation involving the glass-water interface.

Angle of incidence, i = 45°

Angle of refraction = r

From Snell’s law, r can be calculated as:

`(sin "i")/(sin "r") = ""^"w"μ_"g"`

`(sin 45°)/sin "r"` = 1.275

sin r = `(1/(sqrt2))/1.275`

sin r = 0.5546

∴ r = `sin^(-1) (0.5546)` = 33.68°

Hence, the angle of refraction at the water-glass interface is 33.68°.