Light is incident from glass (μ = 1.5) to air. Sketch the variation of the angle of deviation δ with the angle of incident *i* for 0 < *i* < 90°.

#### Solution

Given,

Refractive index of glass, *μ*_{g} = 1.5

Refractive index of air, *μ*_{a}= 1.0

Angle of incidence 0° < *i* < 90

Let us take *θ*_{c} as the Critical angle

\[\Rightarrow \frac{\sin \theta c}{\sin r} = \frac{\mu_a}{\mu_g}\]

\[\Rightarrow \frac{\sin \theta_c}{\sin 90^\circ } = \frac{1}{1 . 5} = 0 . 66\]

\[ \Rightarrow \sin \theta_c = 0 . 66\]

\[ \Rightarrow \theta_c = \sin {}^{- 1} \left( 0 . 66 \right)\]

⇒ *θ*_{c} = 40°48"

The angle of deviation (*δ*) due to refraction from glass to air increases as the angle of incidence increases from 0° to 40°48". The angle of deviation (*δ*) due to total internal reflection further increases from 40°48" to 45° and then it decreases, as shown in the graph.