question_answer

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 \[\frac{1}{\sqrt{3}}m\], 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)  \[{{\tan }^{-1}}\left( \frac{2}{\sqrt{3}-1} \right)\]

Correct Answer: B

Solution :

Here, angle of incidence \[i={{45}^{o}}\] \[=\frac{Lateral\text{ }shift\text{ (}d\text{)}}{Thickness\text{ }of\text{ }glass\text{ }slab\text{ (}t\text{)}~}=\frac{1}{\sqrt{3}}\] 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            \[\frac{d}{t}=\frac{\sin \,{{45}^{o}}\cos r-\cos {{45}^{o}}\sin r}{\cos r}\]                                 \[=\frac{\cos r-\sin r}{\sqrt{2}\,\cos r}\] 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            \[r={{\tan }^{-1}}\left( 1-\frac{\sqrt{2}}{\sqrt{3}} \right)\]