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.

List-I		List-II
A	Angle of reflection	(I)	$2 \sin^{-1} \left( \sqrt{\frac{2}{3}}\right)$
B	Angle of refraction	(II)	$\sin^{-1} \left( \sqrt{\frac{2}{3}} \right) - \sin^{-1} \left( \frac{1}{\sqrt{3}} \right)$
C	Angle between incident and completely polarised rays	(III)	$\sin^{-1} \left( \frac{1}{\sqrt{3}} \right)$
D	Angle of deviation of the incident ray	(IV)	$\cos^{-1} \left( \frac{1}{\sqrt{3}} \right)$

• A. A $\to$ II, B $\to$ III, C $\to$ I , D $\to$ IV.
• B. A $\to$ II, B $\to$ III, C $\to$ IV, D $\to$ I.
• C. A $\to$ IV , B $\to$ I , C $\to$ III, D $\to$ II.
• D. A $\to$ IV , B $\to$ III, C $\to$ I, D $\to$ II.

Answer 1

Answer: (D)

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,
Here, $\sin i_{p} =\mu \sin\, r $
$\mu =1.414=\sqrt{2} $
$\Rightarrow \sin \,r=\frac{\sin \,i_{p}}{\mu}$
$\sin\, r =\frac{1}{\sqrt{2}} \frac{\sqrt{2}}{\sqrt{3}}$
$r =\sin ^{-1} \frac{1}{\sqrt{3}}$
angle of refraction Now, the angle of deviation,
$\delta=i_{p}-r$
$\delta=\sin ^{-1} \sqrt{\frac{2}{3}}-\sin ^{-1} \frac{1}{\sqrt{3}}$
Angle of reflection,
$\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}} $
$\Rightarrow \theta=\cos ^{-1} \frac{1}{\sqrt{3}}$
Angle between incident and completely polarised light is given by
$\phi=i_{p}+\theta$
$\phi=\sin ^{-1} \sqrt{\frac{2}{3}}+\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
$=2 \sin ^{-1} \sqrt{\frac{2}{3}}$