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(B) Given:Thickness $t = 4 \sqrt{3} 	ext{ cm}$Refractive index $\mu = \sqrt{2}$Angle of incidence $i = 	heta_c$ (critical angle)$1$. Calculate the c

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$2$

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(B) Given:Thickness $t = 4 \sqrt{3} 	ext{ cm}$Refractive index $\mu = \sqrt{2}$Angle of incidence $i = 	heta_c$ (critical angle)$1$. Calculate the critical angle:$\sin 	heta_c = \frac{1}{\mu} = \frac{1}{\sqrt{2}}$$	heta_c = 45^{\circ}$So,$i = 45^{\circ}$.$2$. Apply Snell's Law at the first surface:$1 \cdot \sin i = \mu \cdot \sin r$$\sin 45^{\circ} = \sqrt{2} \cdot \sin r$$\frac{1}{\sqrt{2}} = \sqrt{2} \cdot \sin r$$\sin r = \frac{1}{2} \Rightarrow r = 30^{\circ}$.$3$. Calculate lateral displacement $\Delta$:The formula for lateral displacement is $\Delta = \frac{t \sin(i - r)}{\cos r}$.$\Delta = \frac{4 \sqrt{3} \cdot \sin(45^{\circ} - 30^{\circ})}{\cos 30^{\circ}}$$\Delta = \frac{4 \sqrt{3} \cdot \sin 15^{\circ}}{\cos 30^{\circ}}$Given $\sin 15^{\circ} = 0.25 = \frac{1}{4}$ and $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$.$\Delta = \frac{4 \sqrt{3} \cdot (1/4)}{\sqrt{3}/2} = \frac{\sqrt{3}}{\sqrt{3}/2} = 2 	ext{ cm}$.Thus,the lateral displacement is $2 	ext{ cm}$.

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