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 $0<i<{90}^{\circ }$.

Answer 1

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

Given $0^{\circ }<i<90$

Let C $\to$ Critical angle

$\Rightarrow \frac{sinC}{sinr}=-\frac{{\mu }_c}{{\mu }_g}$

$\Rightarrow \frac{sinC}{90}=\frac{1}{15}=0.66$

$\Rightarrow C={40}^{\circ }{48}^{\prime \prime }$

The angle of deviation due to refraction from glass to air increases as the angle of incidence increases from $0^{\circ }$ to ${40}^{\circ }{48}^{\prime \prime }$. The angle of deviation due to total internal reflection further increase for $\left({40}^{\circ }{48}^{\prime \prime }\right)$ to ${45}^{\circ }$ and then it decreases.