A light ray enters through a right-angled prism at point P with an angle of incidence of 45°. It tra

Question

A light ray enters through a right-angled prism at point P with an angle of incidence of 45°. It travels through the prism parallel to its base BC and emerges along the face AC. The refractive index of the prism is

Professor PC Thomas Classes · Answer

(d) : Let the refractive index of the prism be $\mu$. Applying Snell's law at point P: $1 	imes \sin(45^\circ) = \mu \sin(r_1)$, which gives $\sin(r_1) = 1/(\mu\sqrt{2})$. From geometry, $r_1 + r_2 = 90^\circ$, so $r_2 = 90^\circ - r_1$. Applying Snell's law at point Q for grazing emergence (emerges along AC): $\mu \sin(r_2) = 1 	imes \sin(90^\circ)$. This gives $\mu \cos(r_1) = 1$. Using the identity $\cos(r_1) = \sqrt{1 - \sin^2(r_1)}$, we get $\mu \sqrt{1 - 1/(2\mu^2)} = 1$. Squaring both sides yields $\mu^2 (1 - 1/(2\mu^2)) = 1$, so $\mu^2 - 1/2 = 1$, giving $\mu^2 = 3/2$. Thus, $\mu = \sqrt{3/2}$.