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, the angle of refraction produced is

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

Answer 1

Answer: (A)

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

$\frac{\text { Lateral shift }(d)}{\text { Thickness of glass slab }(t)}=\frac{1}{\sqrt{3}}$
Lateral shift $d=\frac{t \sin \delta}{\cos r}=\frac{t \sin (i-r)}{\cos r}$
$\Rightarrow \,\,\,\,\,\frac{d}{t}=\frac{\sin (i-r)}{\cos r}$
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 $\,\,\,\,\, \frac{d}{t}=\frac{1}{\sqrt{2}}(1-\tan r)$
or $\,\,\,\,\, \frac{1}{\sqrt{3}}=\frac{1}{\sqrt{2}}(1-\tan r)$
or $\,\,\,\,\, \tan r=1-\frac{\sqrt{2}}{\sqrt{3}}$
or $\,\,\,\,\, r=\tan ^{-1}\left(1-\frac{\sqrt{2}}{\sqrt{3}}\right)$