Elizabeth

Is it simply a consequence of a reduced duty-cycle resulting from shorter pulses? In other words have you got similar received mean power levels?

Actually, when you are dealing with fs range pulses, I don't think the classical Rayleigh scattering applies, and you need to work in the quantum optics domain.

The fraction of light scattered by a group of scattering particles is the number of particles per unit volume N times the cross-section. For example, the major constituent of the atmosphere, nitrogen, has a Rayleigh cross section of 5.1×10−31 m2 at a wavelength of 532 nm (green light).[8] This means that at atmospheric pressure, where there are about 2×1025 molecules per cubic meter, about a fraction 10−5 of the light will be scattered for every meter of travel.

The strong wavelength dependence of the scattering (~λ−4) means that shorter (blue) wavelengths are scattered more strongly than longer (red) wavelengths.

In optical fibers Rayleigh scattering is an important component of the scattering of optical signals in optical fibers. Silica fibers are glasses, disordered materials with microscopic variations of density and refractive index. These give rise to energy losses due to the scattered light, with the following coefficient

take a look at specific equations at the  following https://en.wikipedia.org/wiki/Rayleigh_scattering  that allow to specify the rigth parameters to measure the scattering for your specific situation.

In fact, you illuminate the sample with broadband radiation λ/dλ=3. Therefore, Raman lines has the same relative width. So broad lines look just as background. 

Elizabeth

Further to the points Richard and Vincent make about signal to noise ratio, does your detection scheme 'gate' the signal, so that your detection window is comparable with the pulse width? If you reduce the detector integration time from 150 fs to 13 fs, you will reduce the volume of liquid from which scattering is detected by a factor 11.5.  This reduction in signal energy will be in addition to any effects arising from changes in the mean power injected.

Hi everyone. Thank you for your answers. I'll try to provide more information.

I am using a photodiode, chopper, and lock-in for my detection. In our old set-up we used a PMT. In the current set-up, from classical Rayleigh scattering cross-section estimates, we should be detecting significantly more signal. BUT, classical Rayleigh scattering does not account for such short pulse length.

As for signal-to-noise, I anticipate higher signal level because I have a higher repetition rate of the laser despite the shorter pulse length.

It makes sense to me that Rayleigh scattering takes some amount of time to occur that depends on the incident intensity and pulse length. I am working to figure out that relationship right now.

Thank you for all of your help!

Sorry to repeat , but if you are dealing with only 3 wavelengths, all of the classical analysis for Rayleigh scattering does not apply.  So to ask "how many cycles does it take to occur" is another way of saying "classical analysis is not valid for 3 wavelengths".  So you need to use quantum optics.

Elizabeth

"I am using a photodiode, chopper, and lock-in for my detection. In our old set-up we used a PMT. In the current set-up, from classical Rayleigh scattering cross-section estimates, we should be detecting significantly more signal."

Are you using a mechanical chopper at a low frequency (e.g. 1 Hz to 1 kHz) compared with the pulse rate?  In the classical case, the scattered signal should depend only on the mean laser input power, the path length of liquid, and the solid angle over which the light is collected. In this limit your signal should be independent of pulse width for the same average input power and wavelength.

Can you estimate the absolute signal power detected in the old experiment?

Are the wavelengths and average laser powers comparable?  Are you collecting a similar proportion of the scattered light?

One significant difference is your use of a photodiode rather than PMT.  A decent large area silicon photodiode will have a noise effective power (NEP) around 10-14 W/sqrt(Hz).  A photomultiplier could easily have a noise floor 100 times smaller.

Could higher noise  from the photodiode explain your problems?  What kind of amplifier do you use between photodiode and lock-in?  What is the chopper frequency?  Noise will increase with frequency when using a large area photodiode with a relatively high junction capacitance ~1 nF.

http://www.hamamatsu.com/eu/en/product/category/3100/3003/3044/H9305-04/index.html

http://www.hamamatsu.com/eu/en/product/category/3100/4001/4103/S1337-1010BR/index.html

Hi everyone,

Thank you for your previous input. It now seems that my question was misguided and it is a problem with the sensitivity of my entire detection scheme.

Thank you for your help.