Yes! Monochromatic Blue Light Can Reflect Red Light

[actionintegral]

Hello everybody,

Is it possible for monochromatic blue light to strike an object and
be reflected as monochromatic red light?

 

[HallsofIvy]

No. monochromatic blue light has a specific wavelength and red light another. If the surface of which the light reflects is painted red, it will absorb the blue light (it looks red under white light because it only reflects red and absorbs other colors) and so will look black.

 

[LURCH]

That is an interesting question. What if the mirror were moving? If the mirror were retreating from the observer, might the light get redshifted? The speed of the lightwaves would still b c for the mirror, but the mirror is in motion wrt the observer, so could the observer shine a blue light on a moving mirror and see red light coming back?

 

[Danger]

Good point. I would expect that to be the case.

 

[cevdet.erturk]

I don't think it's related with motion. Because, it is impossible to move a reflecter with the speed of light's. It's similar to speed of light waves. The answer is neither reasonable nor practical. I think it can be about the structure of the reflecter. Assume a reflecter that can observe all colors apart from red. Maybe it can be a semi-lens. It can make it possible to reflect a red light from a blue one, I think.

 

[cesiumfrog]

Bouncing it off a mirror should doppler shift it, as lurch mentioned. It can also be color shifted if it exchanges momentum with the (very light) mirror.

A better method might be using special crystals like the down-converters used for entanglement experiments, or the frequency doublers applied to monochromatic IR light in monochromatic green laser pointers.

 

[Doc Al]

I don't think it's related with motion. Because, it is impossible to move a reflecter with the speed of light's.

I don't see anything wrong with Lurch's amusing example (although it's probably not the situation the OP had in mind). There's no need for the mirror to be moving at the speed of light, just that the relative speed of the light source and mirror be some significant fraction of light speed. From the mirror frame, the light appears redshifted. The mirror reflects that light, just like any mirror does. The source frame sees that reflected light redshifted again.

 

[LURCH]

From the mirror frame, the light appears redshifted. The mirror reflects that light, just like any mirror does

Good point Doc; hadn't thought of looking at the color shift from the mirror's perspective. So the mirror still sees the light coming at the same speed, but red-shifted when it arrives.

But you are right about this solution being impracticle, cedvet. I mean, maybe one could have the mirror vibrating in place, and time the blue light in pulses that only strike while the mirror is retreating, but I don't think the mirror could be made to vibrate fast enough to get a significant (visible to the naked eye) change in color.

 

[cevdet.erturk]

I don't think the mirror could be made to vibrate fast enough to get a significant (visible to the naked eye) change in color.

What I mean was this. I think we should gain some more information about the structure of monochromatic blue light. Maybe it can make the problem easier.

 

[Danger]

Hey now... I say, old Lurch, I suspect that you might have hit on something there. Hmmm...
I know that there are eavesdropping devices that measure the shift of laser light reflected from windows, which vibrate in response to sounds inside a building. One might be able to build a little desk-top experiment with a cheap laser, a small piece of mirror attached to a piezo crystal, a variable oscillator, and the detector of one's choice. If there is a system to synchronize the laser firing to the stroke cycle of the crystal, you could have constant red or blue shifts occurring. Maybe... hmmm...

 

[LURCH]

Or maybe multiple mirrors, each one shifting the wavelength a bit further into the red...

 

[Danger]

I like that idea on the face of it, but I don't really know anything about mirrors, so how that would be achieved is a mystery to me. I would suspect that your mirror would have to be composed of a material that absorbs at one wavelength and re-emits at a lower one. Am I missing something?

 

[actionintegral]

What if the mechanism of reflection was compton scattering?

 

[lightarrow]

What if the mechanism of reflection was compton scattering?

The mechanism could be right, to decrease a photon's wavelenght, but it's not possible to use it in practice, because the difference in wavelenght from red to blue is too high and it would require a too light particle for the collision:

m = h*(1 - cos(teta))/c*(l_f - l_i)

where l_f and l_i are the wavelenghts after and before the scattering, respectively. So, assuming, for instance:

red light l_f = 600 nm = 600*10^(-9) m and blue light l_i = 500 nm --> l_f - l_i = 100 nm = 10^(-7) m
and 1-cos(teta) = 2:

m = 6.6*10^(-34)/10^(-7)*3*10^8 = 2.2*10^(-35) Kg

that is, about 50,000 times lighter than the electron.

 

[LURCH]

I like that idea on the face of it, but I don't really know anything about mirrors, so how that would be achieved is a mystery to me. I would suspect that your mirror would have to be composed of a material that absorbs at one wavelength and re-emits at a lower one. Am I missing something?

Not the material, but the reflection off of a moving mirror suggested in my earlier post. The light pulse could be timed to strike a mirror while that mirror was retreating, and bounce to another mirror timed so that the pulse arrives when that second mirror is retreating, and so on through a series of mirrors. Theoretically, the light should get shifted slightly toward the red each time it reflects.

 

[lightarrow]

With n reflections on mirrors all of which moves at a speed v, you have:

(l_blue/l_red)^2 = ((c-v)/(c+v))^n

where l_blue is the wavelenght of blue light and l_red the w. of the red.

From this relation you can determine v knowing n or vice versa.

 

[Danger]

Not the material, but the reflection off of a moving mirror suggested in my earlier post.

Okay, thanks for the clarification.

Lightarrow, excellent responses.

 

[lightarrow]

Thank you Danger!