Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

The angle of a prism is $A$ . One of its refracting surfaces is silvered. Light rays falling at an angle of incidence $2A$ on the first surface returns back through the same path after suffering reflection at the silvered surface. The refractive index $\mu$ of the prism is:

$2\sin A$

$2\cos A$

$\frac{1}{2}\cos A$

$\tan A$

Step-by-Step Solution

Condition for Retracing Path: For a light ray to return back through the same path after reflection from a plane mirror (the silvered surface), it must strike the mirror normally. This implies that the angle of incidence at the second surface is zero.
Prism Geometry: For a prism with angle $A$ and refracting angle $r_1$ at the first surface and $r_2$ at the second surface, the relation is $r_1 + r_2 = A$ . Since the ray strikes the second surface normally, $r_2 = 0$ . Therefore, $r_1 = A$ .
Snell's Law at First Surface: Angle of incidence, $i = 2A$ (Given). Angle of refraction, $r_1 = A$ . Using Snell's Law: $\mu_1 \sin i = \mu_2 \sin r_1$ . Assuming the surrounding medium is air ( $\mu_1 = 1$ ), we have: $1 \cdot \sin(2A) = \mu \sin(A)$ .
Trigonometric Identity:

$\sin(2A) = 2 \sin A \cos A$ .

Calculation: $2 \sin A \cos A = \mu \sin A$ $\mu = 2 \cos A$ .

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