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
$\varphi =i_p+\theta$
$\varphi ={\sin }^{-1}\sqrt{\frac{2}{3}}+{\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$
$=2{\sin }^{-1}\sqrt{\frac{2}{3}}$