Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A ray of light is incident at an angle of incidence, $i$ , on one face of a prism of angle $A$ (assumed to be small) and emerges normally from the opposite face. If the refractive index of the prism is $\mu$ , the angle of incidence $i$ , is nearly equal to:

$\mu A$

$\frac{\mu A}{2}$

$\frac{A}{\mu}$

$\frac{A}{2\mu}$

Step-by-Step Solution

Angle of Emergence: The ray emerges normally from the opposite face. This implies that the angle of emergence $e = 0^\circ$ . Consequently, the angle of refraction at the second surface $r_2 = 0^\circ$ .
Prism Relation: For a prism with angle $A$ , the relationship between the refracting angle and the internal refraction angles is $A = r_1 + r_2$ . Since $r_2 = 0$ , we have $r_1 = A$ .
Snell's Law at First Surface: At the first surface, the ray travels from air to the prism. Using Snell's Law: $\sin i = \mu \sin r_1$ .

Substituting $r_1 = A$ , we get $\sin i = \mu \sin A$ .

Small Angle Approximation: The problem states that the prism angle $A$ is small. Consequently, the angle of incidence $i$ will also be small. For small angles (in radians), $\sin \theta \approx \theta$ .

Therefore, $i \approx \mu A$ .

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