Question

The speed of light in media ${M_1}$​ and ${M_2}$ are $1.5 \times {10^8}m{s^{ - 1}}$ and $2 \times {10^8}m{s^{ - 1}}$ respectively. A ray travels from medium ${M_1}$​ to the medium ${M_2}$​ with an angle of incidence $\theta $ . The ray suffers total internal reflection. Then the value of the angle of incidence $\theta $ is.
A) $ > {\sin ^{ - 1}}\left( {\dfrac{3}{4}} \right)$
B) $ < {\sin ^{ - 1}}\left( {\dfrac{3}{4}} \right)$
C) $ = {\sin ^{ - 1}}\left( {\dfrac{3}{4}} \right)$
D) $ \leqslant {\sin ^{ - 1}}\left( {\dfrac{3}{4}} \right)$

Answer 1

Hint:Name the rarer and denser medium. Write down the velocities. As total internal reflection of light is taking place, hence the angle of incidence must be greater than the critical angle. Derive the equation and solve further.

Complete Step by Step solution: Total internal reflection takes place only if the angle of incidence is greater than the critical angle. Lets first write down the given quantities:
Since speed of light in media ${M_1}$ is lesser than speed of light in media ${M_2}$ , therefore media ${M_1}$ is the denser medium and medium ${M_2}$ is the rarer medium.
Now speed of light in medium ${M_1}$ is ${v_d} = 1.5 \times {10^8}m{s^{ - 1}}$
And speed of light in medium ${M_2}$ is ${v_r} = 2 \times {10^8}m{s^{ - 1}}$
Total internal reflection occurs when the light rays travel from a more optically denser medium to a less optically denser medium.
For total internal reflection to occur, the angle of incidence $\theta $ must be greater than the critical angle denoted by ${\theta _c}$ . The required relation is given as:
$\theta > {\theta _c}$
But critical angle is given as:
$ \Rightarrow \theta > {\sin ^{ - 1}}\left( {\dfrac{3}{4}} \right)$
Where ${n_r}$ and ${n_d}$ are the refractive indices of rarer and denser medium respectively.
$ \Rightarrow \theta > {\sin ^{ - 1}}\left( {\dfrac{{{n_r}}}{{{n_d}}}} \right)$
But we have:
$\dfrac{{{n_r}}}{{{n_d}}} = \dfrac{{{v_d}}}{{{v_r}}}$
$ \Rightarrow \theta > {\sin ^{ - 1}}\left( {\dfrac{{{v_d}}}{{{v_r}}}} \right)$
Substituting the values of velocities, we can have
$ \Rightarrow \theta > {\sin ^{ - 1}}\left( {\dfrac{{1.5 \times {{10}^8}}}{{2 \times {{10}^8}}}} \right)$
$ \Rightarrow \theta > {\sin ^{ - 1}}\left( {\dfrac{3}{4}} \right)$
The value of angle of incidence is $\theta > {\sin ^{ - 1}}\left( {\dfrac{3}{4}} \right)$

Therefore, option A is the correct answer.

Additional details: Total internal reflection takes place in diamond, optical fibre cables and mirage.

Note:Remember that total internal reflection only takes place if angle of incidence is greater than the critical angle. Also remember the relation between refractive index and velocity. Be careful while using the comparison sign between critical angle and angle of incidence.