Answer to Question #118782 in Optics for dakota

Question #118782
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?
1
Expert's answer
2020-05-29T09:51:39-0400

Wavelength of light moving in vacuum is λ=550nm=550×109m\lambda=550nm=550\times10^{-9}m .

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

Let, frequency of light is ν\nu and wavelength of light in Diamond is λ1\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

μ=cv\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

μ=νλνλ1    μ=λλ1\mu=\frac{\nu \lambda}{\nu \lambda_1}\\ \implies \mu=\frac{\lambda}{ \lambda_1}\\

Hence, from the above formula we get,


λ1=λμ    λ1=550×1092.42=227.27nm\lambda_1=\frac{\lambda}{\mu}\\ \implies \lambda_1=\frac{550\times 10^{-9}}{2.42}=227.27nm


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

Tau
06.06.20, 13:45

Wonderful explanation!

Leave a comment