JEE Main 2021 · 22 July · evening · Single correct · +4 / −1
A ray of light passes from a denser medium to a rarer medium at an angle of incidence i. The reflected and refracted rays make an angle of 90° with each other. The angle of reflection and refraction are respectively r and r'. The critical angle is given by
- (a) sin⁻¹ (tan r)
- (b) sin⁻¹ (cot r)
- (c) sin⁻¹ (tan r')
- (d) tan⁻¹ (sin i)
Answer and solution
Answer (a) sin⁻¹ (tan r)
Solution
Physics sir · 4 steps
The 90° gap makes r and r' add up to 90°, so Snell's law turns into a cot, and sin C is its reciprocal.
- Angles at the point. The reflected ray is at r from the normal and the refracted ray is at r' on the other side. The angle between them is 180° - r - r' = 90°, so r' = 90° - r. Also r = i by the law of reflection.
- Apply Snell's law. Let μ be the refractive index of the denser medium with respect to the rarer one. Going from denser to rarer, μ sin i = sin r'. Put i = r and sin r' = sin(90° - r) = cos r.
μ = (sin r')/(sin r) = (cos r)/(sin r) = cot r
- Critical angle. For the critical angle C, sin C = 1/μ = 1/(cot r) = tan r.
- So. C = sin⁻¹(tan r), which is option (a).
The idea to keep. When reflected and refracted rays are perpendicular, r + r' = 90° and μ = tan of the incidence angle on the side the light comes from. Then sin C = 1/μ.
Answer from the NTA final answer key. Solution by PrepWiser.