Question

A ray of light is incident on a surface of glass slab at an angle ${45}^o$. If the lateral shift produced per unit thickness is $1/\sqrt{3}$, then 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. ${\tan }^{-1}\left(\sqrt{\frac{2}{3-1}}\right)$

Answer 1

Answer: (B)

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

Lateral shift, $Y=\frac{t\sin (i-r)}{\cos r}$

$\frac{Y}{t}=\frac{\sin (i-r)}{\cos r}$

$=\frac{\sin i\cos r-\cos i\sin r}{\cos r}$

[using identity, $\sin (A-B)=\sin A\cdot \cos B-\cos A\cdot \sin B$]

$=\sin i-\cos i\tan r$

$\Rightarrow \frac{1}{\sqrt{3}}=\sin {45}^o-\cos {45}^o\tan r=\frac{1}{\sqrt{2}}(1-\tan r)$

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

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

$\therefore r={\tan }^{-1}(1-\sqrt{\frac{2}{3}})$.