Question

A ray of light originating in glass is incident on a plane surface separating glass and water. If the speeds of light in glass and water are $2 \times {10^8}m/s$ and $2.25 \times {10^8}m/s$ respectively, find the value of critical angle in glass.

Answer 1

The sine of the critical angle is equal to the reciprocal of the refractive index of the denser medium with respect to the rarer medium. The refractive index of the denser medium with respect to the rarer medium is equal to ratio of the velocity of light in the rarer medium to the velocity of light in the denser medium. Using these formulas and the given values, we can calculate the required answer.

Formula used:
The relation between the refractive index and the critical angle is given by the following expression.
$n = \dfrac{1}{{\sin C}}$
The refractive index can be defined in terms of the velocity of the light in denser medium and the velocity of light in the rarer medium by the following expression.
$n = \dfrac{{{\text{Velocity of light in rarer medium}}}}{{{\text{Velocity of light in denser medium}}}}$

Complete step by step answer:
We are given a ray of light which originates in glass (denser medium) and is incident on a plane surface separating glass and water (rarer medium). The velocity of light in glass and water is given as follows:
${v_w} = 2.25 \times {10^8}m/s \\\ {v_g} = 2 \times {10^8}m/s \\\$
Now from these values of velocity, we can calculate the refractive index of glass with respect to water in the following way.
$n = \dfrac{{{\text{Velocity of light in rarer medium}}}}{{{\text{Velocity of light in denser medium}}}} \\\ = \dfrac{{{\text{Velocity of light in water}}}}{{{\text{Velocity of light in glass}}}} \\\ = \dfrac{{{v_w}}}{{{v_g}}} = \dfrac{{2.25 \times {{10}^8}}}{{2 \times {{10}^8}}} = 1.125 \\\$
Now we know that the sine of the critical angle is equal to the reciprocal of the refractive index of glass with respect to water. Therefore, we can write
$\sin C = \dfrac{1}{n} = \dfrac{1}{{1.125}} = 0.8889 \\\ C = {\sin ^{ - 1}}\left( {0.8889} \right) \\\ C \simeq 62.7^\circ \\\$
This is the required value of the critical angle.

Note:
It should be noted that critical angle is that angle of incidence in the denser medium corresponding to which the angle of refraction in the rarer medium is equal to 90$^\circ$. If the angle of incidence exceeds the critical angle, then we observe total internal reflection instead of refraction.