JEE MAIN - Physics (2022 - 29th July Morning Shift - No. 28)
The X-Y plane be taken as the boundary between two transparent media $$\mathrm{M}_{1}$$ and $$\mathrm{M}_{2}$$. $$\mathrm{M}_{1}$$ in $$Z \geqslant 0$$ has a refractive index of $$\sqrt{2}$$ and $$M_{2}$$ with $$Z<0$$ has a refractive index of $$\sqrt{3}$$. A ray of light travelling in $$\mathrm{M}_{1}$$ along the direction given by the vector $$\overrightarrow{\mathrm{P}}=4 \sqrt{3} \hat{i}-3 \sqrt{3} \hat{j}-5 \hat{k}$$, is incident on the plane of separation. The value of difference between the angle of incident in $$\mathrm{M}_{1}$$ and the angle of refraction in $$\mathrm{M}_{2}$$ will be __________ degree.
Answer
15
Explanation
Normal will be $$ - \widehat k$$ so
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$$\cos i = {{\overleftarrow P \,.\,\widehat n} \over {\left| {\overleftarrow P } \right|\,.\,\left| {\widehat n} \right|}}$$
$${5 \over {10}} = {1 \over 2}$$
$$ \Rightarrow i = 60^\circ $$
and using Snells law
$$\sqrt 2 \sin 60^\circ = \sqrt 3 \sin r$$
$${{\sqrt 3 } \over {\sqrt 2 }} = \sqrt 3 \sin r$$
$$ \Rightarrow r = 45^\circ $$
So, $$i - r = 15^\circ $$
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