The critical angle for light going from medium X into medium Y is $\theta $. The speed of light in medium X is v. The speed of the light in medium Y is
$
  (a){\text{ vcos}}\theta \\
  {\text{(b) }}\dfrac{v}{{\cos \theta }} \\
  (c){\text{ vsin}}\theta \\
  {\text{(d) }}\dfrac{v}{{\sin \theta }} \\
 $

- Hint: In this question use the refractive index of medium X is ${\mu _x}$ and medium Y is ${\mu _y}$. Then use the direct relationship between speed of light, refractive index of that medium and the speed of light that is $v{\mu _x} = C$. Moreover the application of the Snell’s law along with the concept that if the incident ray is incident on a critical angle on one of the medium as shown above then the refracted ray always goes at 90 degree from the normal as shown in the figure. This will help approaching the problem.

Complete step-by-step solution -


Let the refractive index of medium X is ${\mu _x}$ and medium Y is ${\mu _y}$.
Now it is given that the critical angle for light going from medium X into medium Y is $\theta $.
And the speed of the light in medium X is v.
Now we have to find out the speed of light in medium Y.
As we know the multiplication of the speed of light in any medium and the refractive index of that medium is always equal to the speed of light in vacuum.
 So speed of light in medium X is
$ \Rightarrow v{\mu _x} = C$.............. (1)
Let the speed of light in medium Y is ${v_y}$.
So speed of light in medium Y is
$ \Rightarrow {v_y}{\mu _y} = C$.................. (2)
Now from equation (1) and (2) we have,
\[ \Rightarrow {v_y}{\mu _y} = v{\mu _x}\]
\[ \Rightarrow {v_y} = \dfrac{{v{\mu _x}}}{{{\mu _y}}}\]...................... (3)
Now we all know if the incident ray is incident on a critical angle on one of the mediums as shown above then the refracted ray always goes at 90 degree from the normal as shown in the figure.
Now according to Snell’s law we have,
$ \Rightarrow {\mu _x}\sin \theta = {\mu _y}\sin {90^o}$
$ \Rightarrow \dfrac{{{\mu _x}}}{{{\mu _y}}} = \dfrac{{\sin {{90}^o}}}{{\sin \theta }} = \dfrac{1}{{\sin \theta }}$................... (4), $\left[ {\because \sin {{90}^o} = 1} \right]$
Now from equation (3) and (4) we have,
\[ \Rightarrow {v_y} = \dfrac{v}{{\sin \theta }}\]
So this is the required speed of light in medium Y
Hence option (D) is the correct answer.

Note – In general refraction of the light wave moving from one medium into another medium does not happen at any angle, in fact there is a value of angle of incidence of the indenting light at the interface beyond which only the refraction can take place, so the largest angle of incidence above which we can see refraction is called as the critical angle for that incident light ray.