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  • A ray of light is incident on surface of glass slab at an angle . If the lateral shift produced per

A ray of light is incident on surface of glass slab at an angle 45^{\circ}. If the lateral shift produced per unit thickness is \mathrm{\frac{1}{\sqrt{3}}m,} , the angle of refraction produced is:

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer


Here, angle of incidence \mathrm{i=45^{\circ}}
\mathrm{\frac{\text { Lateralshift }(\mathrm{d})}{\text { Thickness of glass slab(t) }}=\frac{1}{\sqrt{3}}} \\ \mathrm{Lateral\ shift, \ d=\frac{t \sin \delta}{\cos r}=\frac{t \sin (i-r)}{\cos r}}
\Rightarrow \frac{\mathrm{d}}{\mathrm{t}}=\frac{\sin (\mathrm{i}-\mathrm{r})}{\cos \mathrm{r}}
\text { or } \quad \frac{\mathrm{d}}{\mathrm{t}}=\frac{\sin \mathrm{i} \cos \mathrm{r}-\cos \sin \mathrm{r}}{\cos \mathrm{r}} \quad \text { or } \quad \frac{\mathrm{d}}{\mathrm{t}}=\frac{\sin 45^{\circ} \cos \mathrm{r}-\cos 45^{\circ} \sin \mathrm{r}}{\cos \mathrm{r}}=\frac{\cos \mathrm{r}-\sin \mathrm{r}}{\sqrt{2} \cos \mathrm{r}}
or \quad \frac{\mathrm{d}}{\mathrm{t}}=\frac{1}{\sqrt{2}}(1-\tan \mathrm{r}) \quad\ or\ \quad \frac{1}{\sqrt{3}}=\frac{1}{\sqrt{2}}(1-\tan r) \quad\ or\ \quad \tan \mathrm{r}=1-\frac{\sqrt{2}}{\sqrt{3}}\\ \mathrm{or\ Angle\ of\ refraction,\ r=\tan ^{-1}\left(1-\frac{\sqrt{2}}{\sqrt{3}}\right)}

Posted by

Suraj Bhandari

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