Question

The velocities of light in 2 different media is $2 \times {10^8}\,m{s^{ - 2}}$ and $3.5 \times {10^8}\,m{s^{ - 2}}$ respectively. The critical angle for these medias is equal to:
A) ${\sin ^{ - 1}}(\dfrac{1}{5})$
B) ${\sin ^{ - 1}}(\dfrac{4}{7})$
C) ${\sin ^{ - 1}}(\dfrac{2}{3})$
D) ${\sin ^{ - 1}}(\dfrac{6}{7})$

Answer 1

Hint
We need to use Snell’s law to find the critical angle. Critical angle is the angle of incidence where the angle of refraction goes to ${90}^{0}$. We will substitute that in Snell’s law.

Complete step-by-step answer
Critical angle between the interfaces of the 2 media is defined as the angle of incidence when the angle of refraction is equal to ${90}^{0}$. Critical angle will only occur when the light travelling from the denser medium to the rarer medium hits the transition between them and is reflected along the transition line.
Using Snell’s law, we have:
${\mu _1}\sin \,i{\text{ }} = {\text{ }}{\mu _2}\sin \,r$
For the case of critical angle i is the critical angle itself and r in 90.
$\sin ({i_c}){\text{ }} = {\text{ }}\dfrac{{{\mu _2}}}{{{\mu _1}}}$
$\mu {\text{ }} = {\text{ }}\dfrac{{speed{\text{ }}of{\text{ }}light{\text{ }}in{\text{ }}vacuum}}{{speed{\text{ }}of{\text{ }}light{\text{ }}in{\text{ }}medium}}$
This means the refractive index of a medium is inversely proportional to the speed of light in the medium,
$\sin ({i_c}){\text{ }} = {\text{ }}\dfrac{{{v_1}}}{{{v_2}}}$
$\sin ({i_c}){\text{ }} = {\text{ }}\dfrac{{2 \times {{10}^8}}}{{3.5 \times {{10}^8}}} \\\ \Rightarrow \sin ({i_c}){\text{ }} = {\text{ }}\dfrac{{20}}{{35}}\, = \,\dfrac{4}{7} \\\ \Rightarrow {i_c} = {\sin ^{ - 1}}(\dfrac{4}{7}) \\\$
Therefore, the option with the correct answer is option (B).

Note
In this question, the speed of light in the medium is given as greater than the speed of light in vacuum which is not possible under any circumstance. But for the question sake, we do consider it.