Question

The velocity of light in a medium is half its velocity in air. If a ray of light emerges from such a medium, into air, the angle of incidence, at which it will be totally internally reflected, is:
A.$15^\circ$
B.$30^\circ$
C.$45^\circ$
D.$60^\circ$

Answer 1

The boundary condition for total internal reflection is that the incident angle should be equal to the critical angle, at which the refracted angle will be $90^\circ$. For solving this question, at first, we may have to find the refractive index of the medium, using the velocity ratio between air and the medium. Then, the obtained refractive index value can be substituted in the Snell’s law to obtain the critical angle.
Formulae used:
${\nu _{med}} = \dfrac{c}{{{n_{med}}}}$
where $c$ is the speed of light in vacuum, ${v_{med}}$ is the speed of light in refractive medium and ${n_{med}}$ is the refractive index of the medium
Snell’s law when applied in the case of total internal reflection, $\dfrac{{{n_{med}}}}{{{n_{air}}}} = \dfrac{{sin90^\circ }}{{sin\theta c}}$
where ${n_{air}}$ is the refractive index of air and $\theta c$ is the critical angle of the medium.

Complete answer:
It is given in the problem that the velocity of light in the medium is half its velocity in air. That is,
${v_{med}} = \dfrac{{{v_{air}}}}{2}$ ......equation(1)
where ${v_{med}}$ and ${v_{air}}$ are the velocities of light in the medium and air, respectively. Now, we can find the refractive index of the medium as:
${n_{med}} = \dfrac{c}{{{v_{med}}}}$ ......equation(2)
Substituting equation (1) in equation(2) we get,
${n_{med}} = \dfrac{c}{{{v_{air}}/2}} = \dfrac{{2c}}{{{v_{air}}}}$
${n_{med}}$ is equal to twice the ${n_{air}}$.
So, we obtained the relation between the two refractive indices as:
${n_{med}} = 2{n_{air}}$
Now, if we recall the Snell’s law, for the case of total internal reflection where the angle of refraction $\theta r = 90^\circ$ and angle of incidence is the critical angle $\left( {\theta c} \right)$, we can write as:
${n_{med}} \times sin\theta c = {n_{air}} \times sin90^\circ$
Using the previous result for the refractive index of the medium, we can write the above equation as:
$2{n_{air}} \times sin\theta c = {n_{air}} \times sin90^\circ$
Therefore, sine of the critical angle will be:
$sin\theta c = \dfrac{{sin90^\circ }}{2} = \dfrac{1}{2}$
Now, we can get the value of the critical angle by finding the sine inverse of $\dfrac{1}{2}$ which will be:
$\theta c = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = 30^\circ$
Therefore, the angle of incidence at which the light will be totally internally reflected is $30^\circ$ .

Hence, the correct answer is option B.

Note:
The velocity of the light in a medium is the ratio of velocity of light in vacuum to the refractive index of the medium. Many of us might have thought that the velocity of light is always a constant everywhere in the universe. But the actual scenario is as per the relation mentioned above. So, we can understand that as the refractive index of the medium changes, the velocity of the light also changes. For example, the refractive index of water is $1.33$ and hence, the light travels $1.33$ times slower in water than in vacuum.