15
$\begingroup$

So here's the setup. We have a spherical source. It emits a pulse of light in all directions with some wavelength $\lambda$. It reflects off of a spherical mirror that is centered around this source.

Now, when the light comes back, it bounces off of the source again (or some percentage, whatever). The source emits another pulse of light at the same time with exactly the same energy as the light that bounced off, but shifted back $\lambda$/$2$, so all the crests line up with all the troughs and the light completely destructively interferes.

I can't see a way out of this. I've spent energy - but where did it go?

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ @Gert (btw thats not what "begs the question" means) Constructive interference on the other side of the source $\endgroup$
    – Señor O
    Commented Mar 14, 2021 at 22:18
  • $\begingroup$ @Gert yes youll frequently seen minor grammar and spelling rules circumvented in online discussion! $\endgroup$
    – Señor O
    Commented Mar 15, 2021 at 4:14
  • 1
    $\begingroup$ @SeñorO Sigh. youll, Senor? (Don't answer that, you'd be wrong) $\endgroup$
    – Gert
    Commented Mar 15, 2021 at 7:58
  • $\begingroup$ This scenario isn't possible. A perfectly reflecting surface such as the one you describe would, by definition, have 0 absorptivity at the wavelength $\lambda$. Zero absorptivity is equivalent to zero emissivity - energy enclosed by the surface cannot escape from it in the form of light with wavelength $\lambda$. It is not a source at all. $\endgroup$
    – Brian Bi
    Commented Mar 15, 2021 at 20:46
  • 1
    $\begingroup$ By the way, you have "completely destructive interference" all the time: All the "paths" that a "ray of light" reaching a specific point didn't take are canceled by interference. (In a sense, a wave spreads from each point in all directions at once; what we describe as directed propagation is the result of all the interferences from all the points.) $\endgroup$
    – Peter - Reinstate Monica
    Commented Mar 16, 2021 at 14:00

3 Answers 3

Reset to default
30
$\begingroup$

The energy went back into your spherical source. For the first pulse the source had to provide energy. For the second pulse the source had to absorb energy. The mistake is simply assuming that the energy required to drive the source is independent of the external fields acting on the source.

Improve this answer
$\endgroup$
4
  • 26
    $\begingroup$ For a more easily visualizable explaination of the last sentence, consider a wave in a pool. If you wish to drive the wave towards a higher energy (constructive interference), you must apply energy into the water. However, if you wish to cause destructive interference, driving the energy of the wave to zero, you actually let the wave push on you, taking its energy, and the effect within the medium is that of "emitting" a pulse of the opposite phase. $\endgroup$
    – Cort Ammon
    Commented Mar 14, 2021 at 23:21
  • $\begingroup$ @CortAmmon Thanks, I didn't understand it until what you said. Much appreciated! $\endgroup$
    – jettae schroff
    Commented Mar 14, 2021 at 23:28
  • 3
    $\begingroup$ @CortAmmon Though with such examples, you always have to be careful not to focus too much on human experience - to a human, it would take effort to cause that destructive interference, even though you're really taking energy from the wave (and mostly dissipating it as heat). It's one of the reasons the concept of physical work is hard to grasp for people without at least some physics background. $\endgroup$
    – Luaan
    Commented Mar 15, 2021 at 12:23
  • 1
    $\begingroup$ @Luaan Very true. I did use an example which includes the human body, and the physics of that is so very tricky! I might have avoided those treacherous issues by transferring the energy of the wave into a spring, and then using a latch to lock the spring back (making it more obvious that energy was taken out of the wave) $\endgroup$
    – Cort Ammon
    Commented Mar 15, 2021 at 15:04
6
$\begingroup$

Think of an analogous but simpler problem. You have a string attached to a wall in one side and you are handling it on the other side (see this video, your hand is the spherical source of light and the wall is the spherical mirror). If you left your hand without moving when the reflection comes back (as in the video) the pulse will bounce again. If you move your hand with the same phase as the reflected pulse you will increase its energy by constructive interference. If you move your hand in anti-phase then the pulse will transmit its energy to your hand and it will disappear because you created the exact opposite wave and canceled it. You can think of this case as moving your hand in such a way as the string "believes" that there is more string to go so the energy keeps flowing from the string into your body. Once it is into your body you do whatever you want with it.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ So, instead of putting power into the antenna, it just impedance-matched it properly and then it received power from the antenna. That's like how RFID chips get their power. $\endgroup$
    – JDługosz
    Commented Mar 15, 2021 at 14:00
2
$\begingroup$

Your question applies perfectly to other waves too, for example, sound wave. The answer is simple: the source does not send out any energy with the "another pulse". It tries to push the media with the "another pulse", but the media moves away by itself, so the source's "push" does not apply any force to the media.

Improve this answer
$\endgroup$
2
  • 2
    $\begingroup$ but there IS no media for the light to go through. With sound, you can probably argue that it turns into heat or something. $\endgroup$
    – jettae schroff
    Commented Mar 14, 2021 at 22:43
  • 1
    $\begingroup$ For light the same principle applies. The source simply does not send out the energy. $\endgroup$
    – verdelite
    Commented Mar 14, 2021 at 23:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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