Unpolarised light from air incidents on the surface of a transparent medium of refractive index 1.414 such that the reflected light is completely polarised. Match the angles given in List-I with the corresponding values given in List-II.
The correct match is Codes $\begin{array}{llll} A & B & C & D \end{array}$

Question

Unpolarised light from air incidents on the surface of a transparent medium of refractive index 1.414 such that the reflected light is completely polarised. Match the angles given in List-I with the corresponding values given in List-II.

The correct match is Codes $\begin{array}{llll} A & B & C & D \end{array}$

Aicharya · Accepted Answer

Figure showing the ray diagram of refraction at a plane surface. Given, $i_p=$ angle of polarisation

So, $ i_p=\tan ^{-1} \mu \Rightarrow i_p=\tan ^{-1} \sqrt{2}=\sin ^{-1} \sqrt{\frac{2}{3}} $ From Snell's law, $ \sin i_p=\mu \sin r $ Here, $ \mu=1.414=\sqrt{2} \Rightarrow \sin r=\frac{\sin i_p}{\mu} $ $ \begin{aligned} \sin r & =\frac{1}{\sqrt{2}} \frac{\sqrt{2}}{\sqrt{3}} \\ r & =\sin ^{-1} \frac{1}{\sqrt{3}} \end{aligned} $ angle of refraction Now, the angle of deviation, $ \begin{aligned} & \delta=i_p-r \\ & \delta=\sin ^{-1} \sqrt{\frac{2}{3}}-\sin ^{-1} \frac{1}{\sqrt{3}} \end{aligned} $ Angle of reflection, $ \begin{gathered} \theta=90^{\circ}-r \Rightarrow r=90^{\circ}-\theta \\ \sin ^{-1} \frac{1}{\sqrt{3}}=90^{\circ}-\theta \Rightarrow \frac{1}{\sqrt{3}}=\sin \left(90^{\circ}-\theta\right) \\ \cos \theta=\frac{1}{\sqrt{3}} \Rightarrow \theta=\cos ^{-1} \frac{1}{\sqrt{3}} \end{gathered} $ Angle between incident and completely polarised light is given by $ \begin{aligned} & \phi=i_p+\theta \\ & \phi=\sin ^{-1} \sqrt{\frac{2}{3}}+\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)=2 \sin ^{-1} \sqrt{\frac{2}{3}} \end{aligned} $ Hence, the correct option is (d)

Unpolarised light from air incidents on the surface of a transparent medium of refractive index 1.414 such…

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