Why does a chopped light beam appear the same colour? Reply from Brad Ferguson, RPI, New York, 21 Dec 2002

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From: "Brad Ferguson"
To: "Derek Abbott"

Say we assume a wavelength of 800 nm (the wavelenght of our laser). This corresponds to a frequency of 375 THz.

Then say we are chopping it at 1 kHz (in our experiment we use 200 Hz). For a square wave at 1 kHz you will have a frequency spectrum with components at odd harmonics ie: 1, 3, 5, 7 kHz etc. The magnitude of each succeeding harmonic decreases linearly so at 21 kHz you have 10 dB less power than in the 1st harmonic, so in practice the square wave still has a finite bandwidth of say 100 kHz.

In the time domain the square wave is multiplied by the sinusoid. Thus in the frequency domain the two signals are convolved and the resulting signal is indeed spread out in frequency. We have components at 375 THz +/- 1, 3, 5, 7 kHz etc. But because 1 kHz is so much less than 1 THz this spread is not visible.

If we increase the chopping rate a lot, say up to 10 THz (this is the chopping frequency required to produce 100 fs pulses ie: our laser) then we see significant frequency broadening of the laser (this is what enables THz to be generated) but the broadening is still not of the order required to change the color (ie 10 THz << 375 THz)

There are still some parts of this I don't understand. ie: if we assume a Gaussian envelope then the 130 fs pulse width of our laser should result in a frequency bandwidth of 7.7 THz but when measured experimentally it is only 1.875 THz.

btw: loved your list of questions. It would take months to try and solve them. Cheers, Brad

A/Prof. Derek Abbott
EEE Dept
University of Adelaide
SA 5005, AUSTRALIA.

dabbott@eleceng.adelaide.edu.au

Ph: +618-8303-5748
Fx: +618-8303-4360

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