Question

Light is incident from glass $(\mu =1.5)$ to air. Sketch the variation of the angle of deviation $\delta$ with the angle of incident $i$ for $0<i<{90}^o$.

Answer 1

Refractive index of glass ${\mu }_g=1.5$

Given, $0^0<i<{90}^0$

Let $C\to$ Critical angle.

$\frac{\sin C}{\sin r}=\frac{{\mu }_a}{{\mu }_g}=\frac{\sin C}{\sin {90}^0}=\frac{1}{15}=0.66$

$C={40}^0{48}^{\prime \prime }$

Th angle of deviation due to refraction from glass to air increases as the angle of icidence increases from $0^0$ to ${40}^{0}48"$. The angle of deviation due to total internal reflection further increases for ${40}^{0}48"$ to ${45}^0$ and then it decreases.