A ray of light is incident at an angle of incidence $$60°$$ on the glass slab of refractive index $$\sqrt{3}$$. After refraction, the light ray emerges out from other parallel faces and lateral shift between incident ray and emergent ray is $$4\sqrt{3}$$ cm. The thickness of the glass slab is ______ cm.

A ray of light is incident at $$60°$$ on a glass slab of refractive index $$\sqrt{3}$$. The lateral shift is $$4\sqrt{3}$$ cm. We need to find the thickness of the glass slab.

Using Snell’s law, we have:

$$n_1 \sin i = n_2 \sin r$$

Substituting the given values yields:

$$1 \times \sin 60° = \sqrt{3} \times \sin r$$

which simplifies to:

$$\frac{\sqrt{3}}{2} = \sqrt{3} \sin r$$

and hence:

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

giving:

$$r = 30°$$

The lateral shift $$d$$ for a parallel glass slab is given by:

$$d = \frac{t}{\cos r} \sin(i - r)$$

Substituting $$i = 60°$$, $$r = 30°$$, and $$d = 4\sqrt{3}$$ cm, we get:

$$4\sqrt{3} = \frac{t}{\cos 30°} \sin(60° - 30°)$$

Since $$\cos 30° = \frac{\sqrt{3}}{2}$$ and $$\sin 30° = \frac{1}{2}$$, this becomes:

$$4\sqrt{3} = \frac{t}{\frac{\sqrt{3}}{2}} \times \frac{1}{2}$$

or equivalently:

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

Multiplying both sides by $$\sqrt{3}$$ gives:

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

Therefore, the answer is $$12$$ cm.