Consider a tank made of glass(refractive index 1.5) with a thick… | JEE Main 2019 (9 Jan, Shift 1)

Q51JEE Main 2019 · 9 JAN · SHIFT 1Difficult

Polarisation

Consider a tank made of glass(refractive index 1.5) with a thick bottom. It is filled with a liquid of refractive index $\mu$ . A student finds that, irrespective of what the incident angle i (see figure) is for a beam of light entering the liquid, the light reflected from the liquid glass interface is never completely polarized. For this to happen, the minimum value of $\mu$ is:

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Answer: (b)

Brewster angle at the liquid–glass surface: $\tan\theta_B = \frac{1.5}{\mu}$ , so $\sin\theta_B = \frac{1.5}{\sqrt{\mu^2 + 2.25}}$ .
Angle in the liquid: $\sin r = \frac{\sin i}{\mu}$ , so the largest possible $r$ has $\sin r_{max} = \frac{1}{\mu}$ .
Never completely polarised means $r$ can never reach $\theta_B$ : $\frac{1}{\mu} < \frac{1.5}{\sqrt{\mu^2+2.25}}$ .
$\mu^2 + 2.25 < 2.25\mu^2 \Rightarrow \mu^2 > \frac{9}{5} \Rightarrow \mu > \frac{3}{\sqrt{5}}$
So the answer is (b) $\frac{3}{\sqrt{5}}$ .

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Polarisation question from JEE Main 2019 (9 Jan, Shift 1)

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