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 $\frac{1}{\sqrt{3}}\text{m}$, the angle of refraction produced is

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

Answer 1

The correct option is B ${\tan }^{-1}(1-\sqrt{\frac{2}{3}})$

Given,

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

and,

$\frac{\text{Lateral shift(d)}}{\text{Thickness of a glass slab}}=\frac{1}{\sqrt{3}}$

Let $r$ be the angle of refraction.

And we know that lateral shift $\left(d\right)$ is

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

$\Rightarrow \frac{d}{t}=\frac{\sin (i-r)}{cosr}$

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

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

Substituting the given values we get,

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

Hence, option (b) is the correct answer.

Why this question? This question provides understanding of application of the formula of lateral shift due to a glass slab.