A monochromatic light of frequency $$5 \times 10^{14}$$ Hz travelling through air, is incident on a medium of refractive index '2'. Wavelength of the refracted light will be:

The frequency of monochromatic light never changes when it passes from one medium to another.
Given frequency $$f = 5 \times 10^{14}\, \text{Hz}$$ remains the same in every medium.

Step 1 : Wavelength in air
Speed of light in air is practically the same as in vacuum, $$c = 3 \times 10^{8}\,\text{m s}^{-1}$$.
Using $$\lambda = \dfrac{v}{f}$$, the wavelength in air is
$$\lambda_{\text{air}} = \frac{c}{f} = \frac{3 \times 10^{8}}{5 \times 10^{14}} = 6 \times 10^{-7}\,\text{m} = 600\,\text{nm}.$$

Step 2 : Speed in the second medium
Refractive index $$\mu$$ of the medium is defined by $$\mu = \frac{c}{v}$$.
For $$\mu = 2$$,
$$v = \frac{c}{\mu} = \frac{3 \times 10^{8}}{2} = 1.5 \times 10^{8}\,\text{m s}^{-1}.$$

Step 3 : Wavelength in the second medium
Again using $$\lambda = \dfrac{v}{f}$$,
$$\lambda_{\text{medium}} = \frac{v}{f} = \frac{1.5 \times 10^{8}}{5 \times 10^{14}} = 3 \times 10^{-7}\,\text{m} = 300\,\text{nm}.$$

Thus the wavelength of the light inside the medium of refractive index $$2$$ is $$300\,\text{nm}$$.

Hence, the correct choice is Option A.