1. Given data

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

2. Find the angle of incidence $i$

For glass to air, critical angle $C$ satisfies

$\sin C = \frac{1}{\mu} = \frac{1}{\sqrt{2}}$

Hence,

$C = 45^\circ$

So,

$i = 45^\circ$

3. Find the angle of refraction inside the glass

Using Snell's law at air-glass interface:

$\sin i = \mu \sin r$

$\sin 45^\circ = \sqrt{2}\sin r$

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

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

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 the values:

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

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

Given:

$\sin 15^\circ = 0.26, \qquad \cos 30^\circ = \frac{\sqrt{3}}{2}$

So,

$d = \sqrt{3}\cdot \frac{0.26}{\sqrt{3}/2}$

$d = \sqrt{3}\cdot 0.26 \cdot \frac{2}{\sqrt{3}}$

$d = 0.52\ \text{cm}$

5. Convert to the required form

The question asks for

$\underline{\hspace{1cm}} \times 10^{-2}\ \text{cm}$

Now,

$0.52\ \text{cm} = 52 \times 10^{-2}\ \text{cm}$

So the required integer is:

$\boxed{52}$