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Question

The angle of a prism is AA. One of its refracting surfaces is silvered. Light rays falling at an angle of incidence 2A2A on the first surface returns back through the same path after suffering reflection at the silvered surface. The refractive index μ\mu of the prism is:

A

2sinA2\sin A

B

2cosA2\cos A

C

12cosA\frac{1}{2}\cos A

D

tanA\tan A

Step-by-Step Solution

  1. 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.
  2. Prism Geometry: For a prism with angle AA and refracting angle r1r_1 at the first surface and r2r_2 at the second surface, the relation is r1+r2=Ar_1 + r_2 = A. Since the ray strikes the second surface normally, r2=0r_2 = 0. Therefore, r1=Ar_1 = A.
  3. Snell's Law at First Surface: Angle of incidence, i=2Ai = 2A (Given). Angle of refraction, r1=Ar_1 = A. Using Snell's Law: μ1sini=μ2sinr1\mu_1 \sin i = \mu_2 \sin r_1. Assuming the surrounding medium is air (μ1=1\mu_1 = 1), we have: 1sin(2A)=μsin(A)1 \cdot \sin(2A) = \mu \sin(A).
  4. Trigonometric Identity:
  • sin(2A)=2sinAcosA\sin(2A) = 2 \sin A \cos A.
  1. Calculation: 2sinAcosA=μsinA2 \sin A \cos A = \mu \sin A μ=2cosA\mu = 2 \cos A.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from RAY OPTICS AND OPTICAL INSTRUMENTS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSRAY OPTICS AND OPTICAL INSTRUMENTSrefractingsurfacessilveredfallingincidence

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