JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Geometrical Optics

A prism of refractive index

μ

and angle of prism

A

is placed in the position of minimum angle of deviation. If minimum angle of deviation is also

A

, then in terms of refractive index value of

A

is :

(A) $2 cos^{- 1} (\frac{μ}{2})$ ✓ Correct
(B) $cos^{- 1} (\frac{μ}{2})$
(C) $sin^{- 1} (\frac{μ}{2})$
(D) $sin^{- 1} (\frac{μ - 1}{2})$

Answer: (A)

Solution

At minimum deviation,

μ = \frac{s i n ( \frac{A + δ _{m}}{2} )}{s i n ( \frac{A}{2} )}

. Given

δ_{m} = A

, we have

μ = \frac{s i n A}{s i n ( A /2 )} = \frac{2 s i n ( A /2 ) c o s ( A /2 )}{s i n ( A /2 )} = 2 cos (A /2)

. Solving for

A

cos (A /2) = \frac{μ}{2}

, so

A = 2 cos^{- 1} (\frac{μ}{2})

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Single correct option

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