Answer to Question #118782 in Optics for dakota

Question

Light of wavelength 550 nm (in a vacuum) passes through a diamond with a refractive index of 2.42.
What is the wavelength of the light while it is in the diamond?

Answer 1

Wavelength of light moving in vacuum is $\lambda=550nm=550\times10^{-9}m$ .

Refractive Index of Diamond is $\mu=2.42$ .

Let, frequency of light is $\nu$ and wavelength of light in Diamond is $\lambda_1$.

Since, refractive index is the ratio of speed of light in vacuum to the speed of light in any other medium,we get

$$\mu=\frac{c}{v}$$

But we also know that frequency of light is independent of medium i.e frequency of light doesn't change, thus

$$\mu=\frac{\nu \lambda}{\nu \lambda_1}\\ \implies \mu=\frac{\lambda}{ \lambda_1}\\$$

Hence, from the above formula we get,

$$\lambda_1=\frac{\lambda}{\mu}\\ \implies \lambda_1=\frac{550\times 10^{-9}}{2.42}=227.27nm$$