Question

A ray of light is incident on a surface of glass slab at an angle $45^\circ . \, $ If the lateral shift produced per unit thickness is $\frac{1}{\sqrt{3 \, }} \, m$ , then the angle of refraction produced at the incident surface is

• A. $\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)$
• B. $\tan ^{-1}\left(1-\sqrt{\frac{2}{3}}\right)$
• C. $\sin ^{-1}\left(1-\sqrt{\frac{2}{3}}\right)$
• D. $\tan ^{-1}\left(\sqrt{\frac{2}{\sqrt{3}-1}}\right)$

Answer 1

Answer: (B)

Here, angle of incidence $i=45^\circ $

$\frac{L a t e r a l \, s h i f t \left(d\right)}{T h i c k n e s s \, o f \, g l a s s \, s l a b \, \left(t\right)}=\frac{1}{\sqrt{3}}$
Lateral shift $d=\frac{t sin \delta}{cos ⁡ r}=\frac{t sin ⁡ \left(\right. i - r \left.\right)}{cos ⁡ r}$
$\Rightarrow \frac{d}{t}=\frac{s i n \left(\right. i - r \left.\right)}{cos ⁡ r}$
or $\frac{d}{t}=\frac{sin i cos ⁡ r - cos ⁡ i sin ⁡ r}{cos ⁡ r}$
or $\frac{d}{t}=\frac{sin 45 ^\circ cos ⁡ r - cos ⁡ 45 ^\circ sin ⁡ r}{cos ⁡ r}=\frac{cos ⁡ r - sin ⁡ r}{\sqrt{2} cos ⁡ r}$
or $\frac{d}{t}=\frac{1}{\sqrt{2}}\left(1 - tan r\right)$
or $\frac{1}{\sqrt{3}}=\frac{1}{\sqrt{2}}\left(1 - tan r\right)$
or $tan r = 1 - \frac{\sqrt{2}}{\sqrt{3}}$
oror $r=\tan ^{-1}\left(1-\frac{\sqrt{2}}{\sqrt{3}}\right)$