A ray of light is incident at an angle of incidence, $i$, on one face of a prism of angle $A$ (assumed to be s

Question

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:

Accepted Answer

1.  **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 $r_2 = 0^\circ$.
2.  **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 $r_1 = A$.
3.  **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 $\sin i = \mu \sin A$.
4.  **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), $\sin 	heta \approx 	heta$.
    *   Therefore, $i \approx \mu A$.

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