A prism of refractive index \mu 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

\sin ^ - 1( \sqrt \mu - 1 \over 2 )

Geometrical Optics PYQ — JEE Main 2021 · 25 Jul · Shift 2 · Q50 with Solution

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Correct answer: A

For a prism in the condition of minimum deviation,

\mu = \frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}

where:

The question says that the minimum angle of deviation is also equal to $A$ . So,

\delta_m = A

Substitute into the formula:

\mu = \frac{\sin\left(\frac{A+A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin A}{\sin\left(\frac{A}{2}\right)}

Using

\sin A = 2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)

we get

\mu = \frac{2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)}

\mu = 2\cos\left(\frac{A}{2}\right)

Thus,

\cos\left(\frac{A}{2}\right) = \frac{\mu}{2}

\frac{A}{2} = \cos^{-1}\left(\frac{\mu}{2}\right)

Therefore,

A = 2\cos^{-1}\left(\frac{\mu}{2}\right)

This corresponds to:

Option A:

2\cos^{-1}\left(\frac{\mu}{2}\right)

So the correct answer is A.