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(B) Given: Refractive index $\mu = \sqrt{3}$,Angle of incidence $i = 60^{\circ}$,Path length inside the slab $L = 5 \, cm$.Using Snell's Law at the fi

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$\frac{5}{2} \, cm$

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(B) Given: Refractive index $\mu = \sqrt{3}$,Angle of incidence $i = 60^{\circ}$,Path length inside the slab $L = 5 \, cm$.Using Snell's Law at the first interface: $\mu_1 \sin i = \mu_2 \sin r_1 \Rightarrow 1 \cdot \sin 60^{\circ} = \sqrt{3} \sin r_1$.$\sin r_1 = \frac{\sin 60^{\circ}}{\sqrt{3}} = \frac{\sqrt{3}/2}{\sqrt{3}} = \frac{1}{2} \Rightarrow r_1 = 30^{\circ}$.The lateral shift $d$ is given by the formula $d = L \sin(i - r_1)$,where $L$ is the distance traveled inside the slab.$d = 5 \sin(60^{\circ} - 30^{\circ}) = 5 \sin 30^{\circ} = 5 	imes \frac{1}{2} = 2.5 \, cm = \frac{5}{2} \, cm$.

$A$ light ray enters a rectangular glass slab of refractive index $\sqrt{3}$ at an angle of incidence of $60^{\circ}$. It travels a distance of $5 \, cm$ inside the slab and emerges out. What is the perpendicular distance between the incident and emergent ray?

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