A monochromatic light wave with wavelength $$\lambda_1$$ and frequency $$\nu_1$$ in air enters another medium. If the angle of incidence and angle of refraction at the interface are 45° and 30° respectively, then the wavelength $$\lambda_2$$ and frequency $$\nu_2$$ of the refracted wave are:

We have angle of incidence = 45° and angle of refraction = 30°. Using Snell's law:

$$\mu = \frac{\sin 45°}{\sin 30°} = \frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}} = \sqrt{2}$$

Now, the frequency of a wave does not change when it passes from one medium to another, so $$\nu_2 = \nu_1$$.

The wavelength in the new medium is related to the wavelength in air by:

$$\lambda_2 = \frac{\lambda_1}{\mu} = \frac{\lambda_1}{\sqrt{2}}$$

Hence, $$\lambda_2 = \frac{1}{\sqrt{2}}\lambda_1$$ and $$\nu_2 = \nu_1$$. So, the correct answer is $$\lambda_2 = \dfrac{1}{\sqrt{2}}\lambda_1, \nu_2 = \nu_1$$.