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

• A

  ${\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right)$
• Correct answer
  B

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

  ${\sin }^{-1}\left(1-\sqrt{\frac{2}{3}}\right)$
• D

  ${\tan }^{-1}\left(\sqrt{\frac{2}{\sqrt{3}-1}}\right)$

Solution

Here, angle of incidence $i=45°$



$\frac{\mathrm{L}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{l}\mathrm{ }\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{f}\mathrm{t}\left(d\right)}{\mathrm{T}\mathrm{h}\mathrm{i}\mathrm{c}\mathrm{k}\mathrm{n}\mathrm{e}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{g}\mathrm{l}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{ }\mathrm{s}\mathrm{l}\mathrm{a}\mathrm{b}\mathrm{ }\left(t\right)}=\frac{1}{\sqrt{3}}$

Lateral shift $d=\frac{t\sin \mathrm{\delta }}{\cos r}=\frac{t\sin (i-r)}{\cos r}$

$\Rightarrow \frac{d}{t}=\frac{\mathrm{s}\mathrm{i}\mathrm{n}(i-r)}{\cos r}$

or $\frac{d}{t}=\frac{\sin i\cos r-\cos i\sin r}{\cos r}$

or $\frac{d}{t}=\frac{\sin 45°\cos r-\cos 45°\sin r}{\cos r}=\frac{\cos r-\sin r}{\sqrt{2}\cos r}$

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 $\tan r=1-\frac{\sqrt{2}}{\sqrt{3}}$

or $r={\tan }^{-1}\left(1-\frac{\sqrt{2}}{\sqrt{3}}\right)$

Hence, the correct answer is option: 'B'.