A light ray is incident on a glass slab of thickness $$4\sqrt{3}$$ cm and refractive index $$\sqrt{2}$$. The angle of incidence is equal to the critical angle for the glass slab with air. The lateral displacement of ray after passing through glass slab is _____ cm. (Given $$\sin 15° = 0.25$$)

The thickness of the glass slab is $$t = 4\sqrt{3}\,\text{cm}$$ and its refractive index with respect to air is $$\mu = \sqrt{2}$$.

Step 1: Find the critical angle
For a glass-air boundary, the critical angle $$c$$ satisfies
$$\sin c = \frac{\mu_{\text{air}}}{\mu_{\text{glass}}} = \frac{1}{\sqrt{2}} \quad -(1)$$
Hence $$c = 45^{\circ}$$ because $$\sin 45^{\circ} = \frac{\sqrt{2}}{2}$$.

The question states that the light ray is incident on the slab at an angle equal to this critical angle, so
$$i = c = 45^{\circ}$$.

Step 2: Use Snell’s law to obtain the refraction angle inside the slab
Snell’s law for the air-glass interface is
$$\mu_{\text{air}}\sin i = \mu_{\text{glass}}\sin r \quad -(2)$$
With $$\mu_{\text{air}} = 1$$ and $$\mu_{\text{glass}} = \sqrt{2}$$, put $$i = 45^{\circ}$$ into $$(2)$$:
$$\sin 45^{\circ} = \sqrt{2}\,\sin r$$
$$\frac{\sqrt{2}}{2} = \sqrt{2}\,\sin r$$
$$\sin r = \frac{1}{2}$$
Therefore $$r = 30^{\circ}$$.

Step 3: Formula for lateral displacement
For a parallel-sided slab, the lateral displacement $$d$$ is
$$d = t\,\frac{\sin (i - r)}{\cos r} \quad -(3)$$

Step 4: Substitute the known values
$$i - r = 45^{\circ} - 30^{\circ} = 15^{\circ}$$
Given $$\sin 15^{\circ} = 0.25$$ and $$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$$, insert all values into $$(3)$$:
$$d = 4\sqrt{3}\;\text{cm}\times \frac{\sin 15^{\circ}}{\cos 30^{\circ}}$$
$$d = 4\sqrt{3}\;\text{cm}\times \frac{0.25}{\frac{\sqrt{3}}{2}}$$
$$d = 4\sqrt{3}\;\text{cm}\times \frac{0.25 \times 2}{\sqrt{3}}$$
$$d = 4\sqrt{3}\;\text{cm}\times \frac{0.5}{\sqrt{3}}$$
$$d = 4 \times 0.5 \;\text{cm} = 2\;\text{cm}$$.

Final Answer:
The lateral displacement of the ray after emerging from the glass slab is $$2\;\text{cm}$$.