Question

Light is incident from glass (μ = 1.50) to water (μ = 1.33). Find the range of the angle of deviation for which there are two angles of incidence.

Answer 1

Given,
Refractive index of glass $:{\mu }_{\mathrm{g}}=1.5=\frac{3}{2}$
Refractive index of water$:{\mu }_{\mathrm{w}}=1.33=\frac{4}{3}$
As per the question,
For two angles of incidence,

1. When light passes straight through the Normal,
⇒ Angle of incidence = 0°
⇒ Angle of refraction = 0°
⇒ Angle of deviation = 0°

2. When light is incident at critical angle θ_c,
$\frac{\sin {\theta }_{\mathrm{c}}}{\sin r}=\frac{{\mu }_{\mathrm{w}}}{{\mu }_{\mathrm{g}}}$
(since the light is passing from glass to water)
$\Rightarrow \sin {\theta }_{\mathrm{c}}=\frac{8}{9}$
$\Rightarrow {\theta }_{\mathrm{c}}={\sin }^{-1}\left(\frac{8}{9}\right)=62.73°$
⇒ Angle of deviation
=90° − θ_c
$=90-{\sin }^{-1}\frac{8}{9}$
$={\cos }^{-1}\frac{8}{9}$
=37.27°

Here, if the angle of incidence increased beyond the critical angle, total internal reflection occurs and deviation decreases.
Therefore, the range of angle of deviation is in between 0 to 37.27° or ${\cos }^{-1}\left(\frac{8}{9}\right)$.