A prism of angle 75° and refractive index √3 is coated with thin film of refractive index 1.5 only at the back exit surface. To have total internal reflection at the back exit surface the incident angle must b $e _______.$
(sin 15° = 0.25 and sin 25° = 0.43)

Question

A prism of angle 75° and refractive index √3 is coated with thin film of refractive index 1.5 only at the back exit surface. To have total internal reflection at the back exit surface the incident angle must b $e _______.$

(sin 15° = 0.25 and sin 25° = 0.43)

G1 Physics · Accepted Answer

At the back surface, light travels from the prism of refractive index √3 to the coating of refractive index 1.5. For total internal reflection, sin c = 1.5/√3 = √3/2. Therefore, c = 60°. Let the refracted angles inside the prism at the first and second surfaces be r₁ and r₂. For a prism, r₁+r₂ = 75°. For total internal reflection at the second surface, r₂ ≥ 60°. Hence, r₁ ≤ 15°. At the first surface, Snell's law gives sin i = √3 sin r₁. Therefore the maximum allowed incident angle corresponds a…

A prism of angle 75° and refractive index √3 is coated with thin film of refractive index 1.5 only at the back exit surface. To have total internal reflection at the back exit surface the incident angle must b e _______. (sin 15° = 0.25 and sin 25° = 0.43)