A ray of light entering from air into a denser medium of refractive index 4/3, as shown in figure. The light ray suffers total internal reflection at the adjacent surface as shown. The maximum value of angle θ should be equal to :

Solution

Physics sir · 4 steps

Bigger θ makes the side-wall angle smaller, so put that angle at the critical angle and work back.

1. Top face. Snell's law at the top surface gives sin θ = 4/3 sin θ'. The normals of the top face and the side face are perpendicular, so the angle of incidence on the side face is θ'' = 90° - θ'.
2. Side face condition. Total internal reflection needs θ'' ≥ C, where sin C = 1/μ = 3/4. As θ grows, θ' grows and θ'' shrinks, so the limiting case is θ'' = C.

   cos θ' = sin θ'' ≥ 3/4 ⇒ sin θ' ≤ √(1-9/16) = √7/4

3. Back into air. Put this into the top-face equation: sin θ ≤ 4/3×√7/4 = √7/3 ≈ 0.88. This is less than 1, so such an angle exists.
4. So. The maximum angle is θ = sin⁻¹√7/3, option (a).

The idea to keep. For a ray entering one face and hitting the adjacent face, the two inside angles add to 90°. Put the critical angle on the second face, then use Snell's law to get back to the outside angle.

Answer from the NTA final answer key. Solution by PrepWiser.