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NEET PhysicsMedium

A ray is incident at an angle of incidence i on one surface of a small angle prism (with angle of prism A) and emerges normally from the opposite surface. If the refractive index of the material of the prism is μ\mu, then the angle of incidence is nearly equal to :

1

A2μ\frac{A}{2\mu}

2

2Aμ\frac{2A}{\mu}

3

μA\mu A

4

μA2\frac{\mu A}{2}

Step-by-Step Solution

Light ray emerges normally, so e=0e = 0. Thus r2=0r_2 = 0. Since r1+r2=Ar_1 + r_2 = A, we have r1=Ar_1 = A. Applying Snell's law on the first surface: 1⋅sin⁡i=μsin⁡r11 \cdot \sin i = \mu \sin r_1. For small angles, sin⁡θ≈θ\sin \theta \approx \theta, so i=μAi = \mu A.

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