Question

When light undergoes refraction at the surface of separation of two media, what happens to its wavelength?

Answer 1

1. The speed of light in a medium$\left(v\right)$ is a product of the frequency$\left(f\right)$ and wavelength of light$\left(\lambda \right)$.
2. Mathematically it is given as
3. $v=f\times \lambda ...\left(1\right)$
4. Now the speed of light depends upon the refractive index$\left(n\right)$ of the medium with the relation $v=\frac{c}{n}...\left(2\right)$, where $c$ is the speed of light.
5. By equating equation $\left(1\right)$and $\left(2\right)$, we get

$$\begin{gathered} f\times \lambda =\frac{c}{n} \\ \lambda =\frac{c}{n\times f} \end{gathered}$$

The frequency of light does not change when light undergoes refraction, i.e. $f$is constant.

Therefore, $\lambda \alpha \frac{1}{n}$

For the optically denser medium, the value of $n$is high and for an optically rarer medium $n$is less.

Hence, the wavelength decreases when light enters a denser medium, and the wavelength increases when light enters a rarer medium.