Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

This question already has an answer here:

Looking at the refractive index of glass, it's around $1.6$.

Then the speed of light $x$ through light should be given by $$ 1.6 = \frac{3.0\times10^8}{x}, $$ so $x$ is about $2\times10^8~\mathrm{m}~\mathrm{s}^{-1}$

The frequency is kept constant, so the wavelength must adapt to suit the slower speed, giving a wavelength of $2/3$ the original.

Does this mean that when passing through glass, say red light (wavelength $650~\mathrm{nm}$) changes to indigo ($445~\mathrm{nm}$), as $650 \times 2/3 = 433~\mathrm{nm}$, or is my logic flawed somewhere?

share|improve this question

marked as duplicate by Chris White, Prahar, Kyle Kanos, Brandon Enright, Danu Jun 29 at 2:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Related: physics.stackexchange.com/q/22385/2451 –  Qmechanic Oct 15 '13 at 16:20

3 Answers 3

up vote 11 down vote accepted

What do you take to define "red" light: a wavelength of $650~\mathrm{nm}$ or a frequency of $460~\mathrm{THz}$? On the one hand, this borders on being an ill-defined question, but I suppose it can be massaged into something answerable.

I would argue that frequency is more fundamental to describing the light. After all, it is the frequency that is constant throughout all this, as you noted. When a photon strikes a receptor in your eye, it doesn't matter whether it did so after just passing through glass or through vacuum - the biochemical response is dictated by the frequency/energy of the photon. Thus it would be more appropriate to say red light stays red, but the wavelength corresponding to red shifts in glass.

share|improve this answer
Hang on, but isn't it wavelength that determines what type of wave an EM wave is? –  DarkLightA Jan 25 '13 at 8:33
@DarkLightA I'm not sure what you mean. At some level, all EM waves are the same. If you're referring to the distinction between UV and infrared, for instance, this is as artificial as color, and may as well be determined by frequency. –  Chris White Jan 25 '13 at 8:36
But on all the images I've seen regarding EM waves like this one: api.ning.com/files/eINH8Km02k3ml9pyPwnYabx*dpD0bCncy8KnwoGsnwPCyV70oTrMQN‌​QX8MSUM0w6a*Ac8UYxZWTh5ptCxKNL1RKPlZ4jYhn6/em_spectrum.jpg it seems to depend on wavelength... –  DarkLightA Jan 25 '13 at 8:40
Well, since we almost always are talking about vacuum or air (index of refraction is very close to 1), it's usually ok to switch between the two and use wavelengths whenever these are better for the situation for one reason or another. Most people can visualize nanometers and kilometers better than megahertz and terahertz, so that's what they use in the diagram. I am still thinking if there are scenarios in which wavelength is fundamentally more important - I may yet add more to the answer. –  Chris White Jan 25 '13 at 8:47
I think Chris has put his finger on the key point. The interation with rhodopsin in your eye depends on the photon energy, and this is unchanged in glass. So if you define "red" as that which excites the red receptors in your eye the colour of the light doesn't change in glass. –  John Rennie Jan 25 '13 at 9:02

I've had the same question some time ago, and discussed it with some friends. First of all: yes, this question is fairly ill-defined, since you only observe color through your eyes, meaning the lightbeam will always be travelling through the eye-fluid when it strikes your photoreceptors, so it's kind of hard to define precisely define the color of light in any other context than when travelling through your eye.

That being said, it turns out what your eyes percieve are the excitations of molecules in your photoreceptors, i.e., the frequency (energy) of the light. Hence, should one try to "translate" the light from the interior of the glass to what your eyes would see, the color would be identical.

share|improve this answer
Huh, does this translate to the Doppler Effect too? Because the frequency is the same and the wavelength is the only thing that changes, does the color we see not change too? –  DarkLightA Jan 25 '13 at 9:09
Hmm? Both frequency and wavelength change due to the Doppler effect, and yes, we do perceive different colors due to Doppler shifts. –  SamRoelants Jan 25 '13 at 9:12

Do not try to characterize light by it's colour. Colour is something that is defined only in vacuum! Light is characterized by it's wavelength or frequency, one is enough. You additionally have to have the knowledge of the refractive index though. So you are right about how the wavelength is changing but the frequency stays constant.

To the question what we will see as an observer: We would see red light. Because when the light quits the glass, the wavelength will experience the reverse change than as when it entered. You would see the exact same light that entered the glass which was described by you as red.

An interesting question though is what would we see if we were able to put our eyes INTO the glass, so that the light is not transformed back to it's initial state. Well then SamRoelants pointed out what we would see.

share|improve this answer
Frequency is unchanged when the wave crosses the air/glass interface, only wavelength and velocity change. –  Thomas Jan 25 '13 at 11:17
@Thomas: Which is what I have written. –  DaP Jan 25 '13 at 11:25
The second paragraph says "wavelength and frequency will experience the reverse change than as when they entered". Of course, this is technically true, since the reverse change of identity is identity, but the way you wrote it suggests frequency has undergone a change. –  Thomas Jan 25 '13 at 11:44
Ok you're right, doesn't sound crystal clear in context. I'll change that. –  DaP Jan 25 '13 at 12:34
I think it is arguable that color is only defined in air. –  Phil Jan 25 '13 at 18:00

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