Given below are two statements : Statement I : If the Brewster's angle for the light propagating from air to glass is \theta_B , then the Brewster's angle for the light propagating from glass to air is \frac\pi2-\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 :

Statement I is true but Statement II is false

Both Statement I and Statement II are false

Both Statement I and Statement II are true

Statement I is false but Statement II is true

Wave Optics PYQ — JEE Main 2023 · 24 Jan · Shift 1 · Q47 with Solution

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Correct answer: D

For light going from medium 1 to medium 2, the Brewster angle $\theta_B$ satisfies

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

where $n_1$ is the refractive index of the incident medium and $n_2$ is that of the refracting medium.

For light propagating from air to glass:

So,

$\tan \theta_B = \mu_g$

Now for light propagating from glass to air:

Let the Brewster angle be $\theta_B'$ . Then

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

But since

$\tan \theta_B = \mu_g,$

we get

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

Hence,

$\theta_B' = \frac{\pi}{2} - \theta_B$

So, Statement I is true.

For light propagating from glass to air, we already found

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

Therefore,

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

But Statement II says

$\theta_B' = \tan^{-1}(\mu_g)$

This is incorrect.

So, Statement II is false.

Therefore, the correct option is:

$\boxed{\text{D}}$