When light waves enter a medium of higher refractive index than the previous, why is it that:

Its wavelength decreases? The frequency of it has to stay the same?

  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/21336 and links therein. $\endgroup$
    – Qmechanic
    Mar 15, 2012 at 0:44
  • $\begingroup$ I think one can explain this by first stating that "the number of waves entering a medium=no. of waves exiting in 1s" Thus the time factor, frequency, must be constant. The only variable thing left is the wavelength, and it has to vary as the speed of light varies, which is due to the EM characteristics (permittivity/permeability) of the material. $\endgroup$ Mar 15, 2012 at 1:06
  • $\begingroup$ But I'm not too sure on this. I think @ColinK could give a better and clearer explanation. $\endgroup$ Mar 15, 2012 at 1:07

4 Answers 4


(This is an intuitive explanation on my part, it may or may not be correct)

Symbols used: $\lambda$ is wavelength, $\nu$ is frequency, $c,v$ are speeds of light in vacuum and in the medium.

Alright. First, we can look at just frequency and determine if frequency should change on passing through a medium.

Frequency can't change

Now, let's take a glass-air interface and pass light through it. (In SI units) In one second, $\nu$ "crest"s will pass through the interface. Now, a crest cannot be distroyed except via interference, so that many crests must exit. Remember, a crest is a zone of maximum amplitude. Since amplitude is related to energy, when there is max amplitude going in, there is max amplitude going out, though the two maxima need not have the same value.

Also, we can directly say that, to conserve energy (which is dependent solely on frequency), the frequency must remain constant.

Speed can change

There doesn't seem to be any reason for the speed to change, as long as the energy associated with unit length of the wave decreases. It's like having a wide pipe with water flowing through it. The speed is slow, but there is a lot of mass being carried through the pipe. If we constrict the pipe, we get a jet of fast water. Here, there is less mass per unit length, but the speed is higher, so the net rate of transfer of mass is the same.

In this case, since $\lambda\nu=v$, and $\nu$ is constant, change of speed requires change of wavelength. This is analogous to the pipe, where increase of speed required decrease of cross-section (alternatively mass per unit length)

Why does it have to change?

Alright. Now we have established that speed can change, lets look at why. Now, an EM wave(like light), carries alternating electric and magnetic fields with it. enter link description here Here's an animation. Now, in any medium, the electric and magnetic fields are altered due to interaction with the medium. Basically, the permittivities/permeabilities change. This means that the light wave is altered in some manner. Since we can't alter frequency, the only thing left is speed/wavelength (and amplitude, but that's not it as we shall see)

Using the relation between light and permittivity/permeability ($\mu_0\varepsilon_0=1/c^2$ and $\mu\varepsilon=1/v^2$), and $\mu=\mu_r\mu_0,\varepsilon=\varepsilon_r\varepsilon_0, n=c/v$ (n is refractive index), we get $n=\sqrt{\mu_r\epsilon_r}$, which explicitly states the relationship between electromagnetic properties of a material and its RI.

Basically, the relation $\mu\varepsilon=1/v^2$ guarantees that the speed of light must change as it passes through a medium, and we get the change in wavelength as a consequence of this.

  • $\begingroup$ The explanation given by Manishearth is excellent. However, the energy will change if you change frequency or wave length, since E=hn or E=hc/l where n and l are the frequency and wave length of light. How to explain this behavior along with the changes in wave length of light when light enter in to a medium? $\endgroup$
    – user11172
    Aug 8, 2012 at 17:00
  • 1
    $\begingroup$ @sst When the wavelength of light enters a medium, it slows down. Therefore the speed is not c anymore. The decrease in velocity is proportional to the decrease in the wavelength, which is how the Energy stays constant in the equation E=hc/l. $\endgroup$
    – Dale
    Dec 4, 2012 at 5:17
  • $\begingroup$ @Manishearth, you state: This means that the light wave is altered in some manner, but it's not clear why it's altered, what makes it altering. $\endgroup$ Feb 5, 2013 at 7:33
  • $\begingroup$ "In this case, since λν=v, and ν is constant, change of speed requires change of wavelength." intuitely, so to say, okay, but not correct, Many related question/answers say that speed of light remains and must remain (intuitively: any velocity below c would hurt rest mass zero). Explanation ist: in certain media light travels longer paths. It is slowed down only virtually as it's got to travel longer routes. Colour, thus wavelenght, does not change. $\endgroup$ Nov 18, 2022 at 13:27

The energy of the light is related to the frequency; when the light enters the medium there are interference patterns that cause the apperent speed of light to change; if the frequency changed, the energy would not be conserved. The wavelength changes to balance the change in speed.

  • $\begingroup$ Try to add some quote from renowned sources for: " if the frequency changed, the energy would not be conserved. The wavelength changes to balance the change in speed." Refering to conservation of energy: if light in fact "lost frequencies" that would heat up the media, so no hurting. $\endgroup$ Nov 18, 2022 at 13:29

Here is a slightly different take on this using the boundary conditions for electromagnetic fields at an interface.

A key boundary condition, that is derived from Faraday's law, is that the component of the E-field tangential to the boundary must be continuous.

So take an EM wave travelling at normal incidence with the electric field solely in a direction tangential to the boundary. Let's represent it as ${\bf E} = E_I \sin (\omega t - kx) \hat{j}$, where I have chosen that the wave travels towards positive $x$ and is polarised in the $y$ direction.

Let the interface be the plane at $x=0$.

The continuity condition then demands that the E-field of the incident wave plus the E-field of the reflected wave must equal the E-field of the transmitted wave, all at $x=0$. This is a condition that must be satisfied for all value of $t$.

Hence $$ E_I \sin (\omega_I t) + E_{R} \sin (\omega_R t) = E_T \sin (\omega_T t)$$

For time-invariant E-field amplitudes, the only way this can be true for all $t$ is if $\omega_I = \omega_R = \omega_T$. i.e. the frequency of the transmitted wave is the same as that of the incident wave. Given that the speed of light in a medium is changed (for reasons explained in Manishearth's answer), then the wavelength of light in the medium must also change.


It's because $v=c/n \equiv \lambda~\nu$, so either $\lambda$ or $\nu$ must change for adapting to new wave momentum as it enters other medium and propagation speed drops. Energy conservation law filters-out change of $\nu$ possibility, because $E=h \nu = \text{const}$ for a photon. So what we left with is $\lambda$.

  • $\begingroup$ Take me for a very bad mathmatician - I thought about your reasoning a lot and came to the conclusion that indeed it's axiomatic: it's given fact, it's a premise, that velocity is constant - c is a constant. Maybe it is the constant of photons the property of which is that they have no "rest mass" (explanation: paths in media are longer). (I learnt here, from you: you take what's been measured and you take as well "what you are left with". $\endgroup$ Nov 18, 2022 at 13:41

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