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Q.
A ray of light is incident normally on one refracting surface of an equilateral prism. If the refractive index of the material of the prism is $1.5,$ then
The ratio of sine of the angle of incidence to sine of angle of refraction $(r)$ is a constant called refractive index.
i.e. $\frac{\sin i}{\sin r}=\mu$ (a constant). For two media, Snell's law can be written as,
${ }_{1} \mu_{2}=\frac{\mu_{2}}{\mu_{1}}=\frac{\sin i}{\sin r}$
Given, refractive index of a prism
$n=1.5=\frac{3}{2}$
As,$\mu=\frac{1}{\sin \theta} \Rightarrow \, \mu=\frac{1}{\sin 60^{\circ}}$
$\mu=\frac{2}{\sqrt{3}} \Rightarrow \sin C=\frac{1}{n}$
$\sin C=\frac{2}{3} \Rightarrow \, C=\sin ^{-1} \frac{2}{3}$
$C=42^{\circ}$
Angle of incidence $>$ critical angle
So, total internal reflection takes places at second refracting surface.