question_answer

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

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

Correct Answer: B

Solution :

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