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. ${\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right)$
• 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 $i=45^{\circ }$

$\frac{\text{ Lateral shift }(d)}{\text{ Thickness of glass slab }(t)}=\frac{1}{\sqrt{3}}$
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$\frac{d}{t}=\frac{\sin 45^{\circ }\cos r-\cos 45^o\sin r}{\cos r}$
$=\frac{\cos r-\sin r}{\sqrt{2}\cos r}$
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 $r={\tan }^{-1}\left(1-\frac{\sqrt{2}}{\sqrt{3}}\right)$