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, μg = 1.5
Refractive index of air, μa= 1.0
Angle of incidence 0° < i < 90
Let us take θc as the Critical angle
\[\Rightarrow \frac{\sin \theta c}{\sin r} = \frac{\mu_a}{\mu_g}\]
\[\Rightarrow \frac{\sin \theta_c}{\sin 90^\circ } = \frac{1}{1 . 5} = 0 . 66\]
\[ \Rightarrow \sin \theta_c = 0 . 66\]
\[ \Rightarrow \theta_c = \sin {}^{- 1} \left( 0 . 66 \right)\]
⇒ θc = 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.
