Consider light travelling from a medium A to medium B separated by a plane interface. If the light undergoes total internal reflection during its travel from medium A to B and the speed of light in media A and B are $$2.4\times10^{8}m/s\text{ and }2.7\times10^{8}m/s$$ respecti vely, then the value of critical angle is :

Refractive index of a medium is defined by the formula $$n = \frac{c}{v}$$, where $$c$$ is the speed of light in vacuum and $$v$$ is the speed of light in the medium.
—(1)

For medium A: $$n_{1} = \frac{c}{2.4\times10^{8}}$$
For medium B: $$n_{2} = \frac{c}{2.7\times10^{8}}$$

Since $$2.4\times10^{8}\;\&lt\;2.7\times10^{8}$$, it follows that $$n_{1}\;\&gt\;n_{2}$$, so total internal reflection is possible when light travels from A to B.

The critical angle $$\theta_{c}$$ is given by Snell’s law at the limit of refraction: $$\sin\theta_{c} = \frac{n_{2}}{n_{1}}$$
—(2)

Substituting the values of $$n_{1}$$ and $$n_{2}$$ from (1), we get
$$\sin\theta_{c} = \frac{\frac{c}{2.7\times10^{8}}}{\frac{c}{2.4\times10^{8}}} = \frac{2.4\times10^{8}}{2.7\times10^{8}} = \frac{8}{9}$$
—(3)

Therefore, $$\theta_{c} = \sin^{-1}\!\Bigl(\frac{8}{9}\Bigr)$$
—(4)

To express $$\theta_{c}$$ in the form involving an inverse tangent, use the identity $$\tan\theta = \frac{\sin\theta}{\sqrt{1-\sin^{2}\theta}}$$.
Substituting $$\sin\theta_{c} = \frac{8}{9}$$ gives
$$\tan\theta_{c} = \frac{\tfrac{8}{9}}{\sqrt{1-\bigl(\tfrac{8}{9}\bigr)^{2}}} = \frac{\tfrac{8}{9}}{\tfrac{\sqrt{17}}{9}} = \frac{8}{\sqrt{17}}$$
—(5)

Hence, $$\theta_{c} = \tan^{-1}\!\Bigl(\frac{8}{\sqrt{17}}\Bigr)\,.\!$$

Comparing with the given options, this corresponds to Option C.