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(B) Given: Angle of incidence $i = 45^{\circ}$.Lateral shift per unit thickness $\frac{d}{t} = \frac{1}{\sqrt{3}}$.The formula for lateral shift $d$ i

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$	an^{-1}\left(1-\sqrt{\frac{2}{3}}ight)$

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(B) Given: Angle of incidence $i = 45^{\circ}$.Lateral shift per unit thickness $\frac{d}{t} = \frac{1}{\sqrt{3}}$.The formula for lateral shift $d$ is given by $d = \frac{t \sin(i - r)}{\cos r}$,where $t$ is the thickness and $r$ is the angle of refraction.Rearranging the formula: $\frac{d}{t} = \frac{\sin(i - r)}{\cos r}$.Expanding $\sin(i - r)$: $\frac{d}{t} = \frac{\sin i \cos r - \cos i \sin r}{\cos r} = \sin i - \cos i 	an r$.Substituting the values: $\frac{1}{\sqrt{3}} = \sin 45^{\circ} - \cos 45^{\circ} 	an r$.Since $\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$,we have: $\frac{1}{\sqrt{3}} = \frac{1}{\sqrt{2}}(1 - 	an r)$.Multiplying by $\sqrt{2}$: $\frac{\sqrt{2}}{\sqrt{3}} = 1 - 	an r$.Therefore,$	an r = 1 - \sqrt{\frac{2}{3}}$.Thus,$r = 	an^{-1}\left(1 - \sqrt{\frac{2}{3}}ight)$.

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

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