1. Key idea: complete polarization on reflection (Brewster condition)

For light incident from a medium of refractive index $n_1$ onto a medium of refractive index $n_2$, the reflected ray is completely polarized only when the angle of incidence at that interface equals the Brewster angle $\theta_B$, where

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

Here, the relevant interface is liquid-glass, with

• liquid refractive index $= \mu$,
• glass refractive index $= 1.5 = \frac{3}{2}$.

So for complete polarization at the liquid-glass interface,

$\tan \theta_B = \frac{3/2}{\mu} = \frac{3}{2\mu}.$

2. What angles can the ray inside the liquid have?

A ray is incident from air on the top surface of the liquid at angle $i$. Let the refracted angle inside the liquid be $r$.

Using Snell's law at the air-liquid surface:

$1\cdot \sin i = \mu \sin r$

$\Rightarrow \sin r = \frac{\sin i}{\mu}.$

As $i$ varies from $0$ to $90^\circ$, the maximum possible value of $\sin i$ is $1$. Hence the maximum possible angle inside the liquid is obtained for grazing incidence:

$\sin r_{\max} = \frac{1}{\mu}$

$r_{\max} = \sin^{-1}\left(\frac{1}{\mu}\right).$

So the angle of incidence at the liquid-glass interface can never exceed $r_{\max}$.

3. Condition for reflected light to never be completely polarized

If complete polarization were possible, then for some incident angle from air, the angle at the liquid-glass interface should become equal to the Brewster angle.

To ensure this never happens, even at the maximum possible internal angle, we need

$r_{\max} < \theta_B.$

For the minimum value of $\mu$ satisfying this, we use the boundary case:

$r_{\max} = \theta_B.$

Thus,

$\sin^{-1}\left(\frac{1}{\mu}\right) = \tan^{-1}\left(\frac{3}{2\mu}\right).$

Take tangent/sine relation by setting the two angles equal. Let this common angle be $\theta$. Then

$\sin \theta = \frac{1}{\mu}, \qquad \tan \theta = \frac{3}{2\mu}.$

Using

$\tan \theta = \frac{\sin \theta}{\cos \theta},$

we get

$\frac{1/\mu}{\cos\theta} = \frac{3}{2\mu}.$

Cancel $1/\mu$:

$\frac{1}{\cos\theta} = \frac{3}{2}$

$\cos\theta = \frac{2}{3}.$

Now,

$\sin\theta = \sqrt{1-\cos^2\theta} = \sqrt{1-\frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt5}{3}.$

But

$\sin\theta = \frac{1}{\mu},$

so

$\frac{1}{\mu} = \frac{\sqrt5}{3}$

$\mu = \frac{3}{\sqrt5}.$

4. Check options

$\mu_{\min} = \frac{3}{\sqrt5}$

This matches Option B.

5. Comparison with stored correct answer

Stored correct answer: B

Our derived answer: B

So the stored answer is correct.