1. Use Snell’s law

At the interface, $n_1 \sin i = n_2 \sin r$

Since the light is coming from air, we take $n_1 \approx 1$. Given: $i=45^\circ, \qquad r=30^\circ$

So, $1\cdot \sin 45^\circ = n_2 \sin 30^\circ$

Substitute values: $\frac{1}{\sqrt{2}} = n_2 \cdot \frac{1}{2}$

Hence, $n_2 = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}} = \sqrt{2}$

2. Relate refractive index to speed

Refractive index of the second medium is $n_2 = \frac{c}{v}$

Thus speed of light in the medium is reduced by a factor $\sqrt{2}$: $v_{\text{medium}} = \frac{c}{\sqrt{2}}$

3. Frequency on refraction

When light passes from one medium to another, its frequency remains unchanged. Therefore, $\nu_2 = \nu_1$

4. Find the new wavelength

Using $\lambda = \frac{v}{\nu}$

In air: $\lambda_1 = \frac{c}{\nu_1}$

In the medium: $\lambda_2 = \frac{c/\sqrt{2}}{\nu_1} = \frac{1}{\sqrt{2}}\lambda_1$

So, $\boxed{\lambda_2 = \frac{1}{\sqrt{2}}\lambda_1, \qquad \nu_2 = \nu_1}$

5. Check options

• A: Wrong, frequency changes.
• B: Wrong, both statements wrong.
• C: Wrong, wavelength should decrease, not increase.
• D: Correct.

Therefore, the correct option is $\boxed{\text{D}}$