The critical angle is the angle of incidence at which light is refracted along the boundary, meaning the angle of refraction is $$90^{\circ}$$. The relationship between the critical angle and the refractive indices of two media can be understood using Snell's Law, given as:

$$ n_1 \sin(\theta_c) = n_2 \sin(90^{\circ}) $$

Here, $$n_1$$ is the refractive index of the first medium, $$n_2$$ is the refractive index of the second medium, and $$\theta_c$$ is the critical angle. Since $$\sin(90^{\circ}) = 1$$, the equation simplifies to:

$$ n_1 \sin(\theta_c) = n_2 $$

Given that the critical angle $$\theta_c$$ is $$45^{\circ}$$, we have:

$$ \sin(45^{\circ}) = \frac{\sqrt{2}}{2} $$

Substituting this value in the simplified Snell's Law equation, we get:

$$ n_1 \left( \frac{\sqrt{2}}{2} \right) = n_2 $$

Therefore,

$$ n_1 = n_2 \cdot \frac{2}{\sqrt{2}} $$

$$ n_1 = n_2 \cdot \sqrt{2} $$

Hence, the ratio of the refractive indices of the first and second media is:

$$ n_1 : n_2 = \sqrt{2} : 1 $$

The correct answer is therefore:

Option D

$$\sqrt{2}: 1$$