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: (a). Perpendicular rays: r + r' = 90°, with i = r. Snell (denser to rarer): μsin r = sin r' = cos r ⇒ tan r = 1/μ. sin C = 1/μ = tan r So the answer is (a) sin ⁻¹(tan r).

A ray of light passes from a denser medium to a rarer medium at an… | JEE Main 2021 (22 Jul, Shift 2)

Q61JEE Main 2021 · 22 JUL · SHIFT 2Medium

Refraction, Glass Slab & Apparent Depth

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 ${}^\circ$ with each other. The angle of reflection and refraction are respectively r and r'. The critical angle is given by

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Answer: (a)

Perpendicular rays: $r + r' = 90^\circ$ , with $i = r$ .
Snell (denser to rarer): $\mu\sin r = \sin r' = \cos r \Rightarrow \tan r = \frac{1}{\mu}$ .
$\sin C = \frac{1}{\mu} = \tan r$
So the answer is (a) $\sin^{-1}(\tan r)$ .

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