A ray of light is incident from air on a glass plate having thickness $$\sqrt{3}$$ cm and refractive index $$\sqrt{2}$$. The angle of incidence of a ray is equal to the critical angle for glass-air interface. The lateral displacement of the ray when it passes through the plate is _____ $$\times 10^{-2}$$ cm. (given sin 15° = 0.26)

Given: Glass plate thickness $$t = \sqrt{3}$$ cm, refractive index $$n = \sqrt{2}$$, angle of incidence = critical angle for glass-air interface.

The critical angle $$i_c$$ satisfies:

$$\sin i_c = \frac{1}{n} = \frac{1}{\sqrt{2}}$$

$$i_c = 45°$$

The angle of incidence $$i = 45°$$. Using Snell's law to find the angle of refraction inside the glass:

$$\sin i = n \sin r$$

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

$$\sin r = \frac{1}{2} \implies r = 30°$$

The lateral displacement formula for a parallel-sided glass slab is:

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

Substituting the values:

$$d = \frac{\sqrt{3} \times \sin(45° - 30°)}{\cos 30°} = \frac{\sqrt{3} \times \sin 15°}{\cos 30°}$$

Using $$\sin 15° = 0.26$$ and $$\cos 30° = \frac{\sqrt{3}}{2}$$:

$$d = \frac{\sqrt{3} \times 0.26}{\frac{\sqrt{3}}{2}} = \frac{0.26 \times 2}{1} = 0.52 \text{ cm}$$

Converting to the required form:

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

Therefore, the answer is $$\mathbf{52}$$.