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Example situation

If I knew that this green light, which has a wavelength of 500 nm in air, travels through a medium with a refractive index of 2.4, how would I find the wavelength in the medium?

These are the formulas that I do know:

$$\lambda_0=\frac{1}{f}V_0$$

$$ c=\lambda f $$

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

But I do not understand how all of these equations would play out in finding out the wavelength?

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  • $\begingroup$ In this problem, you can assume that air has a refractive index of 1. And you should know that $f$ doesn't change when the light enters the medium. So working out $\lambda$ in the medium is just simple algebra. $\endgroup$
    – PM 2Ring
    Jul 21, 2018 at 0:46
  • $\begingroup$ @PM2Ring I was wondering are there any other equations i should be aware of? because i know only one eqn that has refractive index inside of it- $n*sin(\theta)=n*sin(\theta)$ but it doesn't apply to this problem though $\endgroup$
    – John Rawls
    Jul 21, 2018 at 0:53
  • $\begingroup$ No other equations are required. Note that your 2nd equation is just a version of your 1st equation for the case where $v=c$, slightly rearranged. $\endgroup$
    – PM 2Ring
    Jul 21, 2018 at 1:05

1 Answer 1

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The frequency always remain the same in all media. So the speed of light and the wavelength change. The refractive index tells you how much the speed of light changes: $$ v = \frac{c}{n} , $$ where $n$ is the refractive index, $c$ is the speed of light in vacuum and $v$ is the speed of light in the medium.

In vacuum, the speed of light, wavelength $\lambda$ and frequency $\nu$ are related by $$ c = \lambda \nu , $$ and in the medium, it is related by $$ v = \lambda_m \nu . $$ So now one can substitute these two equations into the first equation, cancel off the frequency and get $$ \lambda_m = \frac{\lambda}{n} . $$

Physically, one can understand it as follows. In the medium, light propagates slower. Therefore, the wavelength would needs to be shorter.

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