Question

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°.

Answer 1

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
$\Rightarrow \frac{\sin \theta \mathrm{c}}{\sin r}=\frac{{\mu }_{\mathrm{a}}}{{\mu }_{\mathrm{g}}}$
$$\begin{gathered} \Rightarrow \frac{\sin {\theta }_{\mathrm{c}}}{\sin 90°}=\frac{1}{1.5}=0.66 \\ \Rightarrow \sin {\theta }_{\mathrm{c}}=0.66 \\ \Rightarrow {\theta }_{\mathrm{c}}=\sin {}^{-1}\left(0.66\right) \end{gathered}$$
⇒ 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.