Understanding Total Internal Reflection (TIR): For light to remain within the rod, it must undergo total internal reflection at the lateral face. TIR occurs when light travels from a denser medium (higher refractive index) to a rarer medium (lower refractive index) at an angle greater than the critical angle.

Critical Angle ($$\theta_c$$): The critical angle is the angle of incidence at which the angle of refraction is 90 degrees. It's given by:

$$\sin(\theta_c) = \frac{1}{n}$$

where $$n$$ is the refractive index of the rod.

Snell's Law at the Entry Point: Let $$\theta_i$$ be the angle of incidence at the point where light enters the rod, and $$\theta_r$$ be the angle of refraction inside the rod. According to Snell's Law:

$$1 \cdot \sin(\theta_i) = n \cdot \sin(\theta_r)$$

Here, the refractive index of air is approximately 1.

Geometry Inside the Rod: The angle of incidence at the lateral face is related to $$\theta_r$$. Let's call the angle of incidence at the lateral face $$\theta$$. From geometry, we can deduce that:

$$\theta = 90^\circ - \theta_r$$

Condition for TIR at the Lateral Face: For TIR to occur at the lateral face, $$\theta$$ must be greater than the critical angle $$\theta_c$$:

$$\theta > \theta_c$$

$$90^\circ - \theta_r > \theta_c$$

$$\sin(90^\circ - \theta_r) > \sin(\theta_c)$$

$$\cos(\theta_r) > \frac{1}{n}$$

Combining Equations: We want to find the minimum value of $$n$$ such that light will not leave the rod for any value of $$\theta_i$$. The maximum value of $$\theta_r$$ occurs when $$\theta_i = 90^\circ$$.

From Snell's Law:

$$\sin(90^\circ) = n \cdot \sin(\theta_r)$$

$$1 = n \cdot \sin(\theta_r)$$

$$\sin(\theta_r) = \frac{1}{n}$$

Thus, $$\theta_r = \sin^{-1}\left(\frac{1}{n}\right)$$.

Applying the Condition:

We have $$\cos(\theta_r) > \frac{1}{n}$$. We also know that $$\sin(\theta_r) = \frac{1}{n}$$. Using the identity $$\sin^2(\theta_r) + \cos^2(\theta_r) = 1$$, we get:

$$\cos(\theta_r) = \sqrt{1 - \sin^2(\theta_r)} = \sqrt{1 - \frac{1}{n^2}}$$

So, we need:

$$\sqrt{1 - \frac{1}{n^2}} > \frac{1}{n}$$

Squaring both sides:

$$1 - \frac{1}{n^2} > \frac{1}{n^2}$$

$$1 > \frac{2}{n^2}$$

$$n^2 > 2$$

$$n > \sqrt{2}$$

Therefore, the least value of $$n$$ for which light, once entered, will not leave the rod through its lateral face is $$\sqrt{2}$$.

So, the correct answer is:

Option A: $$\sqrt{2}$$

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