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## 23 March 2011

### Mail Time: Is Color Determined by Wavelength or Frequency?

 Visible spectrum from betesoft.com
Darren Tarshis, a physics teacher in Hayward, CA, has some physical optics questions:

Imagine that red light with a wavelength of 600 nm passes from air to a chunk of, say, diamond. In the diamond, I know the speed slows, which causes the wavelength to shorten (because the frequency remains constant). In the diamond, would the light have a different color because of its new wavelength?

I always teach my students that for a sound wave, pitch is determined by the wavelength/frequency, and for a light wave, color is determined by the wavelength/frequency. But I'm starting to think this may be incorrect, and the pitch is actually determined by frequency only, not wavelength, and color is determined by wavelength only, not frequency.

Yup. Frequency determines color and pitch.  The red light stays red even in diamond.

As a quick example: My voice is baritone. Imagine that you are in the pool with your ears just under water, and I am standing on deck talking to you. When the sound waves from my voice enter the water, they start moving about 4 times faster. The frequency doesn't change -- frequency of a wave NEVER changes when the wave changes materials -- so the wavelength increases by a factor of four as well. If pitch were determined by wavelength, then my voice would sound not only soprano, but squeaky soprano.

Similarly, have you ever stood underwater and looked up at the trees overhanging the pool?  The leaves of the trees still look green, even though the light speed (and thus the wavelength) has decreased by 25%.

I also propose a fanciful biological rationale for pitch being related to frequency only. The eardrum vibrates in response to incoming sound waves. It is the rate of vibration -- the frequency -- that can be measured by the ear and converted to a frequency. But how would an ear measure wavelength? With a meterstick? With a teeny weeny tape measure that an invisible goblin sticks out of the ear to measure the peak-to-peak distance of the incoming sound wave?

As long as the sound wave is in room temperature air, or as long as the light wave is in a vacuum (or air), then wavelength and frequency can be used to desribe color and pitch interchangably. That's why it's perfectly okay to say that red light is about 700 nm, and violet is about 400 nm. Those wavelength values must change when the light enters diamond, but the frequency of a given color will never change.

GCJ

1. and for light the energy of a photon is hf, independent of the material it's in. The rods and cones in your eye are responding to that energy being deposited.

2. In the case of sound, the perceived frequency is the frequncy of vibration of the eardrum, so in water the frequency i sthe same, wavelength different. For the eye, the rhodpsin is in your retinal cells, which are in vitreous humor - so the medium never changes. The wavelength for any frequency would be the same no matter what medium it enters your eyes from. And for light, it is accepted practice to describe light by it's wavelength in a vacuum (c/f) rather than it's frequency. Maybe beacuse 100's of nm are easier to work with than 100's of THz.
Also, remember that photons don't actually slow down in a medium, they are absorbed and re-emitted, so the *average* speed slows, but the instantaneous speed of a photon is always c. So the actual wavelength of a photon of a given frequency never changes!
So for sound, yes, for light, it doesn't make a difference.

3. Ed, you're correct about light. No arguments there. However, an introductory physics student in the physical optics unit should be considering light as a wave whose frequency indicates color. The details you've provided will blow the minds of most general or AP level juniors and seniors. The discussion in the post about frequency determining color will allow introductory students to solve any problem they might be faced with. Then when they're in their sophomore optics class, they can learn the details you've described.

4. Thank you for this post. I am teaching AP Physics for the first time this year and was just wondering about this very same question. Now I know!

5. Thank you for this post. I am teaching AP Physics for the first time this year and was just wondering about this very same questions. Now I know the answer!

6. Remember that velocity (v) is frequency (f) times wavelength (w). In a medium other than vacuum, the wavelength is w=w'/n where w' is the vacuum wavelength and the velocity is v=c/n. Then E=hf becomes E=hv/w=h(c/n)/(w'/n)=hc/w'. Anytime you see a spectrum shown as a function of wavelength, the vacuum wavelength is implied.