JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Geometrical Optics

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

9 0^{\circ}

with each other. The angle of reflection and refraction are respectively

r

and

r^{'}

. The critical angle is given by :

Answer: (D)

Solution

The reflected ray makes the angle

r = i

with the normal on the incident side and the refracted ray makes

r^{'}

with the normal on the far side, so the angle between them is

18 0^{\circ} - r - r^{'}

. Setting this equal to

9 0^{\circ}

gives

r^{'} = 9 0^{\circ} - r

.
Snell's law at the boundary, with

n_{1}

for the denser medium and

n_{2}

for the rarer one, gives

n_{1} sin i = n_{2} sin r^{'} \Rightarrow n_{1} sin r = n_{2} cos r

\frac{n _{2}}{n _{1}} = tan r

The critical angle

C

for this pair of media satisfies

sin C = \frac{n _{2}}{n _{1}}

, so

C = sin^{- 1} (tan r)

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Single correct option