# WaveLengths of Light Waves and their respective photons

1. May 8, 2014

### user3

Suppose you have a source of light that emits light with a wavelength of 2 meters, and you set the device to be turned on and switched off alternately. You also set it so that each interval the device is turned on is only long enough for 1 meter to be emitted (1/2 a wavelength). Do you ever observe any photons?

2. May 8, 2014

### UltrafastPED

Since photons can be modelled as point particles ... the answer is yes.

And if your detector provides an energy spectrum, it will always give the energy for one of these photons.

3. May 8, 2014

### sugeet

This question is absurd, if the photons emitted have wavelength of 2 meters, then we assume that source can emit only 2meters(wavelegth) photons. Its not like in shorter time shorter wavelengths come out and in longer time longer one`s come out, radiation is discreet!

We must also understand how photons are produced, they are generally through de-excitation of electrons in energy levels, however only visible photons can be produced that way, for wavelengths of 2 meters(VHF-range), they use other methods, but photons are discreet and the wavelength output does not depend on the on/off time. The delta-t, will affect only the spectral width.

I hope it helped.

4. May 8, 2014

### user3

How does the alternate switching affect photon density? If it has no effect, can it be used to reduce power consumption at transmission antennas?

Last edited: May 8, 2014
5. May 8, 2014

### user3

I did not say shorter time intervals produce shorter wavelengths. Instead, I said it they will produce fractions of a wavelength.

6. May 9, 2014

### UltrafastPED

For a wavelength of λ the time for a complete cycle is Δt=λ/c=1/frequency.

Your device has a shutter that cycles open-close twice during Δt.

You must first check your assumptions-start with classical electromagnetic radiation: if the wavelength/frequency is fixed then the emitter must be running continuously - you understand why this is true?

Your "chopper" will then increase the bandwidth of the system - this is a consequence of the time-bandwidth theorem, which is a very general result for classical waves. The details of how it does so varies with the physical setup, but it always takes place.

As you increase the speed of the chopper the bandwidth continues to increase; if the emitter is inside a cavity (enclosed space) the radiation that does not escape will accumulate ("rattle around") inside the cavity. Eventually it will reach some equilibrium state - and your device will approach that of a black body.

You can actually build a small radio transmitter (or buy one) and test this. If it was a laser experiment I would use an electrooptic shutter; you will have to find something else for radio waves.

With this classical description in mind, a quantum optical model can be built.

What should we expect? For a system which is periodically excited and then emits a single photon - I think your detector will sometimes click, and sometimes not.

7. May 11, 2014

### Robert_G

In my opinion, this depends on what kind of device you are using to make system shut down and switch on. if this device is in the laser cavity, well, your laser-system must very efficient to observe the light, because the photon only passes the gain medium once and then should be blocked by your device. If it is put outside the cavity, you device is just a chopper. and yes, you will observe the laser coming out, just the number of the photon is small, and the bandwidth is very large.

8. May 12, 2014

### DrDu

Last edited: May 12, 2014
9. May 12, 2014

### user3

It's the first time that I have been told that. How is that? What about this E = h f ?