JEE MAIN - Physics (2021 - 22th July Evening Shift - No. 19)
A ray of light passing through a prism ($$\mu$$ = $$\sqrt 3 $$) suffers minimum deviation. It is found that the angle of incidence is double the angle of refraction within the prism. Then, the angle of prism is _____________ (in degrees).
Answer
60
Explanation
For minimum deviation r1 = r2 = A/2
given i = 2r
$$\mu = {{\sin i} \over {\sin r}} = {{\sin 2r} \over {\sin r}}$$
$$ \Rightarrow \cos r = {\mu \over 2}$$
$$\Rightarrow$$ r = 30$$^\circ$$
$$\Rightarrow$$ A = 60$$^\circ$$
given i = 2r
$$\mu = {{\sin i} \over {\sin r}} = {{\sin 2r} \over {\sin r}}$$
$$ \Rightarrow \cos r = {\mu \over 2}$$
$$\Rightarrow$$ r = 30$$^\circ$$
$$\Rightarrow$$ A = 60$$^\circ$$
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