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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Reflection at the liquid-glass interface is fully polarized when the angle in the liquid equals Brewster's angle \(\tan\theta_B=1.5/\mu\). Compare it with the largest angle possible inside the liquid, \(\sin^{-1}(1/\mu)\).
Light enters the liquid from air with \(i\) from \(0\) to \(90^\circ\), so the angle of refraction \(r\) in the liquid satisfies \(\sin r=\frac{\sin i}{\mu}\le\frac1\mu\); the maximum \(r_{max}=\sin^{-1}\frac1\mu\), with \(\tan r_{max}=\frac{1}{\sqrt{\mu^2-1}}\). At the liquid-glass interface, reflected light is completely polarized if \(r=\theta_B\) where \(\tan\theta_B=\frac{1.5}{\mu}\). It is never completely polarized if \(\theta_B\gt r_{max}\): \(\frac{1.5}{\mu}\gt\frac{1}{\sqrt{\mu^2-1}}\Rightarrow 2.25(\mu^2-1)\gt\mu^2\Rightarrow 1.25\mu^2\gt2.25\Rightarrow\mu^2\gt\frac95\). So \(\mu\gt\frac{3}{\sqrt5}\); the minimum (limiting) value is \(\frac{3}{\sqrt5}\). Answer: (2) \(\frac{3}{\sqrt{5}}\)
