question_answer

A ray of light is incident on a surface of glass slab at an angle 45\[^{0}C\] . 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)  \[{{\tan }^{-1}}\left( 1-\frac{\sqrt{2}}{3} \right)\]

C) \[{{\sin }^{-1}}\left( 1-\frac{\sqrt{2}}{3} \right)\]                              

D) \[ta{{n}^{-1}}\left( \sqrt{\frac{2}{\sqrt{3}-1}} \right)\]

Correct Answer: B

Solution :

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