1. Given data

• Thickness of slab: $t = 4\sqrt{3}\ \text{cm}$
• Refractive index of glass: $\mu = \sqrt{2}$
• Angle of incidence is equal to the critical angle for glass-air.

2. Find the angle of incidence

For glass to air, critical angle $C$ satisfies: $\sin C = \frac{1}{\mu} = \frac{1}{\sqrt{2}}$

So, $C = 45^\circ$

Hence, $i = 45^\circ$

3. Find the angle of refraction inside the slab

Using Snell’s law at air-glass interface: $\sin i = \mu \sin r$

Substitute $i=45^\circ$ and $\mu=\sqrt{2}$: $\frac{1}{\sqrt{2}} = \sqrt{2}\sin r$

$\sin r = \frac{1}{2}$

Therefore, $r = 30^\circ$

4. Formula for lateral displacement

For a parallel-sided glass slab, lateral displacement is: $d = t\,\frac{\sin(i-r)}{\cos r}$

Substitute values: $d = 4\sqrt{3}\cdot \frac{\sin(45^\circ-30^\circ)}{\cos 30^\circ}$

$d = 4\sqrt{3}\cdot \frac{\sin 15^\circ}{\cos 30^\circ}$

Given: $\sin 15^\circ = 0.25$ and $\cos 30^\circ = \frac{\sqrt{3}}{2}$

So, $d = 4\sqrt{3}\cdot \frac{0.25}{\sqrt{3}/2}$

$d = 4\sqrt{3}\cdot 0.25\cdot \frac{2}{\sqrt{3}}$

$d = 2$

5. Final answer

The lateral displacement is: $\boxed{2\ \text{cm}}$

This matches the stored correct answer.