Given below are two statements :
Statement I : If the Brewster's angle for the light propagating from air to glass is $$\theta_B$$, then Brewster's angle for the light propagating from glass to air is $$\frac{\pi}{2} - \theta_B$$.
Statement II : The Brewster's angle for the light propagating from glass to air is $$\tan^{-1}\mu_g$$, where $$\mu_g$$ is the refractive index of glass.
In the light of the above statements, choose the correct answer from the options given below :

We are given two statements about Brewster's angle and need to verify their truth.

First, recall the formula for Brewster's angle. When light travels from a medium with refractive index $$n_1$$ to a medium with refractive index $$n_2$$, Brewster's angle $$\theta_B$$ is given by:

$$\tan \theta_B = \frac{n_2}{n_1}$$

Now, analyze Statement I: If Brewster's angle for light propagating from air to glass is $$\theta_B$$, then Brewster's angle for light propagating from glass to air is $$\frac{\pi}{2} - \theta_B$$.

For light propagating from air to glass:

Refractive index of air, $$n_1 \approx 1$$, and refractive index of glass, $$n_2 = \mu_g$$.

So, Brewster's angle $$\theta_B$$ satisfies:

$$\tan \theta_B = \frac{\mu_g}{1} = \mu_g$$

Thus, $$\theta_B = \tan^{-1}(\mu_g)$$.

For light propagating from glass to air:

Refractive index of glass, $$n_1 = \mu_g$$, and refractive index of air, $$n_2 \approx 1$$.

Let Brewster's angle be $$\theta_B'$$. Then:

$$\tan \theta_B' = \frac{n_2}{n_1} = \frac{1}{\mu_g}$$

Thus, $$\theta_B' = \tan^{-1}\left(\frac{1}{\mu_g}\right)$$.

Now, consider $$\frac{\pi}{2} - \theta_B$$:

$$\tan\left(\frac{\pi}{2} - \theta_B\right) = \cot \theta_B = \frac{1}{\tan \theta_B} = \frac{1}{\mu_g}$$

Therefore, $$\frac{\pi}{2} - \theta_B = \tan^{-1}\left(\frac{1}{\mu_g}\right)$$.

This matches $$\theta_B'$$, so Statement I is true.

Now, Statement II: The Brewster's angle for light propagating from glass to air is $$\tan^{-1}(\mu_g)$$.

From above, Brewster's angle for glass to air is $$\theta_B' = \tan^{-1}\left(\frac{1}{\mu_g}\right)$$, not $$\tan^{-1}(\mu_g)$$.

Note that $$\tan^{-1}(\mu_g)$$ is $$\theta_B$$, the angle for air to glass, not glass to air.

Thus, Statement II is false.

Therefore, Statement I is true, but Statement II is false.

The correct option is B.