Consider a tank made of glass (refractive index 1.5 ) with a thick bottom. It is filled with a liquid of refractive index μ . A student finds that, irrespective of what the incident angle i (see figur

Question

Consider a tank made of glass (refractive index $1.5$ ) with a thick bottom. It is filled with a liquid of refractive index $μ .$ 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 $μ$ is:

TestHub · Accepted Answer

For reflected light not be completely polarized,

$\tan r \neq \frac{μ_{g}}{μ_{w}} = \frac{1.5}{μ}$

For minimum $μ r$ is maximum. $r$ is maximum when $i$ is maximum i.e. $i = 90^{o}$

$\Rightarrow \sin r = \frac{1}{μ}$

$\Rightarrow \tan r = \frac{1}{\sqrt{μ^{2} - 1}}$

$∴ \frac{1}{\sqrt{μ^{2} - 1}} = \frac{1.5}{μ}$

$\Rightarrow μ^{2} = 2.25 (μ^{2} - 1)$

$\Rightarrow 1.25 μ^{2} = 2.25$

$\Rightarrow μ = \sqrt{\frac{2.25}{1.25}}$

$= \sqrt{\frac{9}{5}} = \frac{3}{\sqrt{5}}$

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