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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 $\frac{1}{\sqrt{3}} \mathrm{~m}$, the angle of refraction produced is
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The correct answer is:
$\tan ^{-1}\left(1-\sqrt{\frac{2}{3}}\right)$
Here, angle of incidence $i=45^{\circ}$
$\frac{\text { Lateral shift }(\mathrm{d})}{\text { Thickness of glass slab (t) }}=\frac{1}{\sqrt{3}}$
Lateral shift $\mathrm{d}=\frac{\mathrm{t} \sin \delta}{\cos \mathrm{r}}=\frac{\mathrm{t} \sin (\mathrm{i}-\mathrm{r})}{\cos \mathrm{r}}$
$\Rightarrow \quad \frac{\mathrm{d}}{\mathrm{t}}=\frac{\sin (\mathrm{i}-\mathrm{r})}{\cos \mathrm{r}}$
or $\quad \frac{\mathrm{d}}{\mathrm{t}}=\frac{\sin \mathrm{i} \cos \mathrm{r}-\cos \mathrm{i} \sin \mathrm{r}}{\cos \mathrm{r}}$
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 $\quad \frac{\mathrm{d}}{\mathrm{t}}=\frac{1}{\sqrt{2}}(1-\tan \mathrm{r})$
or $\quad \frac{1}{\sqrt{3}}=\frac{1}{\sqrt{2}}(1-\tan \mathrm{r})$
or $\quad \tan \mathrm{r}=1-\frac{\sqrt{2}}{\sqrt{3}}$
or $\quad \tan { }^{-1}\left(1-\frac{\sqrt{2}}{\sqrt{3}}\right)$
$\frac{\text { Lateral shift }(\mathrm{d})}{\text { Thickness of glass slab (t) }}=\frac{1}{\sqrt{3}}$
Lateral shift $\mathrm{d}=\frac{\mathrm{t} \sin \delta}{\cos \mathrm{r}}=\frac{\mathrm{t} \sin (\mathrm{i}-\mathrm{r})}{\cos \mathrm{r}}$
$\Rightarrow \quad \frac{\mathrm{d}}{\mathrm{t}}=\frac{\sin (\mathrm{i}-\mathrm{r})}{\cos \mathrm{r}}$
or $\quad \frac{\mathrm{d}}{\mathrm{t}}=\frac{\sin \mathrm{i} \cos \mathrm{r}-\cos \mathrm{i} \sin \mathrm{r}}{\cos \mathrm{r}}$
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 $\quad \frac{\mathrm{d}}{\mathrm{t}}=\frac{1}{\sqrt{2}}(1-\tan \mathrm{r})$
or $\quad \frac{1}{\sqrt{3}}=\frac{1}{\sqrt{2}}(1-\tan \mathrm{r})$
or $\quad \tan \mathrm{r}=1-\frac{\sqrt{2}}{\sqrt{3}}$
or $\quad \tan { }^{-1}\left(1-\frac{\sqrt{2}}{\sqrt{3}}\right)$
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