Given data

• Angle of incidence: $i = 60^\circ$
• Refractive index of slab: $\mu = \sqrt{3}$
• Lateral shift: $d = 4\sqrt{3}\,\text{cm}$

Find the angle of refraction using Snell's law

Snell's law: $\mu = \frac{\sin i}{\sin r}$

So, $\sqrt{3} = \frac{\sin 60^\circ}{\sin r}$

Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$

therefore, $\sqrt{3} = \frac{\sqrt{3}/2}{\sin r}$

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

Hence, $r = 30^\circ$

Use formula for lateral shift in a parallel-sided glass slab

The lateral shift is $d = t\,\frac{\sin(i-r)}{\cos r}$

where $t$ is the thickness of the slab.

Substitute $i=60^\circ$, $r=30^\circ$: $d = t\,\frac{\sin(60^\circ-30^\circ)}{\cos 30^\circ}$

$d = t\,\frac{\sin 30^\circ}{\cos 30^\circ}$

$d = t\,\frac{1/2}{\sqrt{3}/2} = \frac{t}{\sqrt{3}}$

Substitute the given lateral shift

$4\sqrt{3} = \frac{t}{\sqrt{3}}$

$t = 4\sqrt{3}\times \sqrt{3} = 4 \times 3 = 12\,\text{cm}$

Final answer $\boxed{12}$

Comparison with stored answer

Stored correct answer: $12$

My derived answer is also $12$, so they agree.