5
$\begingroup$

There are many threads on this topic (like this one) but one aspect about the equation $ v = \lambda\nu $ still confuses me.

I have read that frequency does not change when light crosses into different media. But since light traveling in any media has a slower speed than light traveling in a vacuum, clearly the wavelength $ \lambda $ must be reduced in order to satisfy the above equation.

My question is: how can frequency remain constant while wavelength decreases?

The equation $ v = \lambda \nu $ suggests that wavelength and frequency are inversely proportional, so if one decreases, the other should increase. But when light crosses into a new medium, if wavelength changes but frequency doesn't, that seems to imply that the wavelength of light and frequency of light are unrelated.

$\endgroup$
  • $\begingroup$ I think this is a very good question. The frequency of a wave is set by the frequency of the excitation source. When the wave encounters a discontinuity in refractive index the solutions on both sides of the discontinuity have to be in phase at all times (except for a constant phase shift at that boundary). This can only be satisfied if the frequencies of both parts of the wave are the same. $\endgroup$ – CuriousOne Dec 23 '14 at 2:21
  • $\begingroup$ Thank you! So if I am following correctly, the gist is that an "external factor" (frequency of the excitation source), that is outside the scope of this equation, is what "sets" the frequency of the light wave.Would it be accurate to say that the if there frequency were to change, at an interface between media or otherwise, the wave would have to experience some kind of interference? And that this is not something that an interface between media provides? (Only interacting with another light wave would result in interference?) $\endgroup$ – user46549 Dec 23 '14 at 3:24
  • $\begingroup$ The gist is that there is a phase continuity condition at the plane where the medium changes. One can express all of this with interfering waves, too, but those need to have the same frequency to satisfy the proper interference, just as well. $\endgroup$ – CuriousOne Dec 23 '14 at 4:47
  • $\begingroup$ Related, though not really a duplicate: Why frequency doesn't change during refraction? $\endgroup$ – John Rennie Dec 23 '14 at 7:13
1
$\begingroup$

I may not be understanding the source of your difficulty. There are three facts here.

First, the speed, frequency and wavelength are related as $v = \lambda \nu$.

Second, the frequency of light remains the same when crossing the interfaces between media. This is a consequence of ensuring that the continuity conditions implied by Maxwell's equations are satisfied across the interface. If there were a frequency change, there could be no fixed phase relationship between time-varying fields either side of the boundary, so (for example) the tangential components of the E-fields and H-fields could not be continuous. If this is your problem, then your question is a duplicate of Why doesn't the frequency of light change during refraction?

Third, the speed of light is slower, by a factor $\sqrt{\mu_r \epsilon_r}$, in a material with relative permeabilities or permittivities greater than unity. Again, this is a simple consequence of solving Maxwell's equations in media with finite polarisation and/or magnetisation. See for example Why does larger permittivity of a medium cause light to propagate slower?

If you accept these three facts then the behaviour of light as it crosses from one medium to another is not mysterious. The speed is not constant, therefore there isn't an inverse relation between the frequency and wavelength. The frequency is constant, therefore the speed and wavelength are proportional.

$\endgroup$
0
$\begingroup$

If speed decreases at constant frequency that means obligatorily that the wavelength must diminuish. Your circular reasoning does not consider that the equation's inverse proportionality depends on λ.

Example: If frequency is 1 MHz (1.000.000/sec) and wavelength 1 μm, the velocity is 1m/sec. If wavelength decreases to 0,5 μm, velocity is diminuishing to 0,5 m/sec.In both cases your equation is fulfilled. In the second case the curve is simply flatter than in the first case."Wave length" is equivalent with "speed".

$\endgroup$
-1
$\begingroup$

The relation velocity equal $ v = \lambda\nu $ is a natural relation. As frequency is number of waves passing a point per second and Wavelength is length of each wave so on multiplying these you get the distance covered i.e. velocity. And It is rule of nature to keep frequency of light constant on changing medium(ans for 'why' I believe it is not in scope of your syllabus book otherwise you may not be asking such question). And when you said as one decrease other should increase you assume velocity to be constant and that's not the case.

$\endgroup$
  • 2
    $\begingroup$ Thank you as well, it does seem like I was assuming that velocity was held constant when I said that. I suppose that at least for the basic level that I am at, the "hold velocity constant" assumption should be made when comparing different frequencies of light in the same medium (e.g. among electromagnetic waves traveling in a vacuum, gamma rays have high frequency and short wavelength while radio waves have low frequency and long wavelength) while the "hold frequency constant" assumption should be made when considering one light wave crossing into a different medium. $\endgroup$ – user46549 Dec 23 '14 at 3:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.