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 & Solution

Correct Answer

(B) \(\frac{3}{{\sqrt 5 }}\)

Detailed explanation

\(\sin 90^{\circ}=\mu \sin \theta\)

\(\Rightarrow \quad \sin \theta=\frac{1}{\mu}\)

\({\mu \sin \theta=1.5 \sin r}\)

\({\mu \tan \theta=1.5}\)

\( \Rightarrow \tan \theta = \frac{{1.5}}{\mu }\)

\(\sin \theta = \frac{3}{{\sqrt {9 + 4{\mu ^2}} }} = \frac{1}{\mu }\)

\(9{\mu ^2} = 9 + 4{\mu ^2}\)

\( \Rightarrow \mu = \frac{3}{{\sqrt 5 }}\)

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