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For a prism kept in air, it is found that for an angle of incidence $60 °$ , the angle of refraction ' $A$ ', angle of deviation ' $δ$ ' and angle of emergence 'e' become equal. The minimum angle of incidence of a ray that will be transmitted through the prism is

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1.73

1.15

1.5

1.33

answer is A.

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Detailed Solution

Given i = 60°, A = $δ$ = e

$δ$ = i +e - A $\Rightarrow$ $δ$ = i

( $∵$ e = A) and $δ$ = i = e

$μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin \frac{A}{2}}$

Here, angle of deviation is minimum ( $∵$ i = e)

$μ = \frac{\sin (\frac{60 ° + 60 °}{2})}{\sin (60 ° / 2)} = \sqrt{3}$

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