Question

A ray of light is incident at 32 degree on a glass ice surface .After refraction it is deviated by 3 degree . Calculate the critical angle for ice ,if the critical angle for glass is 42 degree . (Given n glass > n ice )

Answer 1

Critical angle for ice is 46.5°

Explanation:

If the refractive index of ice w.r.t. glass is $\mu_{i/g}$

Then, by Snell's law

$\mu_{i/g}=\frac{\sin i}{\sin r}$

Given that

$i=32^\circ$

The ray is deviated by 3° and since ice is rarer medium than glass, therefore, the deviation will be away from normal

$r=32^\circ+3^\circ=35^\circ$

Therefore,

$\mu_{i/g}=\frac{\sin 32^\circ}{\sin 35^\circ}$

If the refractive index of air w.r.t. glass be $\mu_{a/g}$ and refractive index of air w.r.t. ice be $\mu_{a/i}$  then

$\mu_{a/g}=\sin 42^\circ$

$\implies \mu_{g/a}=\frac{1}{\sin 42^\circ}$

If the critical angle of ice is $\theta$

Then

$\mu_{a/i}=\sin\theta$

$\implies \mu_{i/a}=\frac{1}{\sin\theta}$

Again

$\mu_{i/g}=\frac{\mu_{i/a}}{\mu_{g/a}}$

$\implies \frac{\sin 32^\circ}{\sin 35^\circ}=\frac{1/\sin\theta}{1/\sin 42^\circ}$

$\implies \frac{\sin 32^\circ}{\sin 35^\circ}=\frac{\sin 42^\circ}{\sin\theta}$

$\implies \sin\theta=\frac{\sin 42^\circ \times\sin 35^\circ}{\sin 32^\circ}$

$\implies \sin\theta=\frac{0.669\times 0.574}{0.53}$

$\implies\sin\theta=0.7245$

$\implies \theta=\sin^{-1}0.7245$

$\implies \theta=46.43^\circ$

Hope this answer is helpful.

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