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
\[: \mu_g = 1 . 5 = \frac{3}{2}\]
Refractive index of water \[: \mu_g = 1 . 5 = \frac{3}{2}\]
Refractive index of water \[:    \mu_w  = 1 . 33 = \frac{4}{3}\]
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, 
\[\frac{\sin \theta_c}{\sin r} = \frac{\mu_w}{\mu_g}\](since the light is passing from glass to water)
\[\Rightarrow   \sin   \theta_c  = \frac{8}{9}\]
\[\Rightarrow \theta_c = \sin^{- 1} \left( \frac{8}{9} \right) = 62 . 73^\circ \]
⇒ Angle of deviation
=90° − θ_c
\[= 90 - \sin^{- 1} \frac{8}{9}\] = 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  \[\cos^{- 1} \left( \frac{8}{9} \right)\]