Light travels in two media $\mathrm{M}_{1}$ and $\mathrm{M}_{2}$ with speeds $1.5 \times 10^{8} \mathrm{~ms}^{-1}$ and $2.0 \times 10^{8} \mathrm{~ms}^{-1}$ respectively. The critical angle between them is:

(A) $\tan ^{-1}\left(\frac{3}{\sqrt{7}}\right)$

(B) $\tan ^{-1}\left(\frac{2}{3}\right)$

(C) $\cos ^{-1}\left(\frac{3}{4}\right)$

(D) $\sin ^{-1}\left(\frac{2}{3}\right)$

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Geometrical Optics

Light travels in two media

M_{1}

and

M_{2}

with speeds

1.5 \times 1 0^{8} ms^{- 1}

and

2.0 \times 1 0^{8} ms^{- 1}

respectively. The critical angle between them is:

Answer: (A)

The refractive indices are inversely proportional to speeds:

n \propto \frac{1}{v}

. For critical angle from denser to rarer medium:

sin i_{c} = \frac{n _{r}}{n _{d}} = \frac{v _{d}}{v _{r}} = \frac{1.5 \times 1 0 ^{8}}{2.0 \times 1 0 ^{8}} = \frac{3}{4}

. Therefore,

cos i_{c} = 1 - sin^{2} i_{c} = 1 - \frac{9}{16} = \frac{7}{4}

, giving

tan i_{c} = \frac{s i n i _{c}}{c o s i _{c}} = \frac{3/4}{7 /4} = \frac{3}{7}

. Thus,

i_{c} = tan^{- 1} (\frac{3}{7})

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Single correct option