JEE Main 2022 · 29 June · evening · Single correct · +4 / −1
The speed of light in media 'A' and 'B' are 2.0 × 10¹⁰ cm/s and 1.5 × 10¹⁰ cm/s respectively. A ray of light enters from the medium B to A at an incident angle 'θ'. If the ray suffers total internal reflection, then
- (a) θ = sin ⁻¹( 3/4 )
- (b) θ > sin ⁻¹( 2/3 )
- (c) θ < sin ⁻¹( 3/4 )
- (d) θ > sin ⁻¹( 3/4 )
Answer and solution
Answer (d) θ > sin ⁻¹( 3/4 )
Solution
Physics sir · 4 steps
Light goes from slower B into faster A, so find the critical angle and make θ bigger than it.
- Which side is denser. Refractive index is n = c/v, so the medium where light is slower is optically denser. Light is slower in B, so B is denser, and TIR is possible for a ray going from B into A.
- Critical angle. At the critical angle C the refracted ray grazes the surface, so n_B sin C = n_A sin 90°. Since n = c/v, the ratio of indices is the inverse ratio of speeds.
sin C = n_A/n_B = v_B/v_A = (1.5 × 10¹⁰)/(2.0 × 10¹⁰) = 3/4
- Condition for TIR. Total internal reflection happens only when the angle of incidence is greater than the critical angle, so θ > C = sin⁻¹(3/4).
- So. The ray is totally internally reflected when θ > sin⁻¹(3/4), which is option (d).
The idea to keep. For TIR, light must go from the slower medium to the faster one, and sin C = v_(denser)/vᵣₐᵣₑᵣ. Then the angle of incidence must be strictly greater than C.
Answer from the NTA final answer key. Solution by PrepWiser.