# Need of LED Modulation

I have an experiment of calculating the velocity of light by using phase difference produced (link of the experiment is at the end). The experiment has a control unit which has an LED installed to produce light. The information on the experiment manual says A high intensity and high frequency LED is modulated at 50.1MHz by a crystal controlled oscillator..... My question is what is the need to modulate an LED which is already producing light of high frequency and high intensity?

https://drive.google.com/open?id=1WUDU0kmoFolZHg77SNLWmuFbgdO8-WZq The above line mentioned is on the very first page.

• This question would be better in Electrical Engineering, in my opinion. – Casimir Rönnlöf Nov 8 '18 at 13:38
• Minor comment to the post (v2): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. – Qmechanic Nov 8 '18 at 13:50
• @CasimirRönnlöf - Disagree. It's an engineering project, sure. But the question is a physics question, not an engineering question. – Samuel Weir Nov 8 '18 at 16:40
• If you assume that light is a wave phenomenon with a "frequency," then you can use an interferometer to measure the phase difference between two different, unmodulated light rays emitted by the same, coherent source. But I think the point of modulating the beam is to eliminate any assumption about the nature of light from the experiment. You'll know for a fact that there is a modulation wave of known frequency if you build the modulator. – Solomon Slow Nov 8 '18 at 16:55

The proposed method is based on the comparison between the phases of the transmitted and received light. More specifically, it comes down to finding the round trip distance the light has to travel to cause a $$180^{\circ}$$ phase shift of the modulating frequency.

First, an electrical signal with frequency $$50.1$$MHz is applied to the LED and modulates the intensity of the LED light. After the round trip, the light is received by a photo sensor, which converts it back to the electrical signal. All further processing is done in the electrical, not optical, domain, i.e., we are comparing the phases of two $$50.1$$MHz electrical signals, transmitted and received.

Since the period of $$50.1$$MHz signal is $$19.96$$ns and half period is $$9.98$$ns, the round trip distance has to be about $$0.3m/ns\times10ns=3m$$, which corresponds to a one way trip of $$150$$cm, which conveniently close to the length of the base plate scale.

So, we can see that the frequency of the modulating signal was specifically selected to maximize the usage of the available base plate length.

The frequency of a red LED light is on the order of $$500$$THz, which would be impossible to process in the electrical domain.

Hopefully, this makes it clear why, for this test, we need to modulate the LED light with a relatively low frequency electrical signal.

• Hi,..thanks a lot, that was really a deep one, Hats Off.. – Onkar Singh Nov 12 '18 at 4:47

When you transmit an unmodulated carrier you do not know how much time is spent between one specific part of the carrier as it is transmitted and when it is received. Modulation is a form of labeling the carrier cycles so when you send and receive them you know which one corresponds to which one. Otherwise you will have millions and billions of zero crossings or EM peaks, say, and how do you know which is which. The instructions describe a classic technique to measure time of arrival by first modulating that is labeling portions of a carrier and then upon reflection measuring the time of arrival of the modulation. Here it is intensity modulation used that in RF is called amplitude modulation.