Question: 45: Two transparent media A and B are separated by a plane boundary. The speed of light in those media are 1.5 \times 10^{8} \mathrm{~m} / \mathrm{s} and 2.0 \times 10^{8} \mathrm{~m} / \mathrm{s}, respectively. The critical angle for a ray of light for these two media is
(1) \tan ^{-1}(0.500)
(2) \tan ^{-1}(0.750)
(3) \sin ^{-1}(0.500)
(4) \sin ^{-1}(0.750)
Answer: Option (4)
Explanation:
The refractive index of a medium is given by n = \frac{c}{v},
where c is the speed of light in vacuum and v
is the speed of light in the medium.
Let medium A have speed v_1 = 1.5 \times 10^{8} \mathrm{~m\,s^{-1}}
and medium B have speed v_2 = 2.0 \times 10^{8} \mathrm{~m\,s^{-1}}.
Since light travels slower in medium A, it is optically denser.
The refractive indices are proportional to the inverse of speeds,
so \frac{n_2}{n_1} = \frac{v_1}{v_2}.
The critical angle C for light going from denser to rarer medium is given by
\sin C = \frac{n_2}{n_1}.
Substituting the values,
\sin C = \frac{1.5 \times 10^{8}}{2.0 \times 10^{8}} = 0.75.
Therefore, the critical angle is \sin^{-1}(0.75).
Hence, the correct answer is Option (4).