How Can Pulse Duration Change After Passing Through a Material?

[Voxynn]

I have a laser with a short pulse duration (sub 20 fs) and i want to know how long the pulse would be after it passes through X amount of material. I don't know the refractive index of the material, but I do know its GDD over the wavelength I'm operating at.

How do I go about calculating the new pulse duration?

Thanks.

 

[gomboc]

If you know the group delay dispersion, it's easy to relate this to the total dispersion coefficient D_\lambda (which is simpler to work with in this problem, in my opinion):

GDD = \frac{-2\pi c}{\lambda^2}D_\lambda

Once you know the dispersion coefficient, the problem is a fairly simple group delay problem - if your initial pulse is 20 fs, you can find the spectral width \Delta\lambda of your pulse:

\Delta\lambda = \frac{0.441\lambda_0^2}{c\Delta t}

where \lambda_0 is the average wavelength of your pulse. (The 0.441 comes from Fourier analysis and uncertainty principle, assuming a Gaussian peak - chances are you have this formula somewhere)

Now finishing up is easy. Using the two values you've just found, the spread in group delay \Delta\tau_g is simply:

\Delta\tau_g = LD_\lambda (\Delta\lambda)

The pulse will always lengthen, so your new pulse length is the original length plus the calculated group delay spread:

\Delta t_{new} = 20\ fs\ +\Delta\tau_g

Hope that helps. You might have been given a simpler way to do that problem using GDD directly, but if there is such a method I'm not yet aware of it!

 

[Voxynn]

Thanks Gomboc.

One question though: dispersion coefficient can also be found through the equation

D(lam)= -lam/c * d2n/dlam2

where d2n is the second derivative of the material's refractive index against wavelength.
Since the second deriviative of MOST glasses & crystals is positive, the material dispersion coefficient is negative and so

broadening = D(lam)*Bandwidth

is also negative. Doesn't this suggest a SHORTENING of the pulse?

Sorry for the confusion!

 

[gomboc]

Hmm. Well, the value of D_\lambda is a continuously defined function of wavelength for all materials relative to refractive index, and most materials (if not all) have a wavelength where the group velocity reaches a minimum, and at that point the dispersion coefficient becomes positive - i.e. for glass, D_\lambda is greater than zero for \lambda > 1.31\ \mu m.

The sign of D_\lambda just indicates what type of dispersion takes place. For a negative value, the shorter wavelengths in a given pulse travel slower than the longer wavelengths, meaning the short wavelengths are the last to arrive. For a positive value, the opposite is true.

To conceptualize it, I just imagine the pulse as a ball of uniformly distributed red and blue specks (the uniform distribution helps me to remember that despite the pulse's \Delta\lambda, they all still enter the material at precisely the same time. Then, suppose you throw the ball through the air, knowing that the red specks will always travel slightly faster than the blue specks. When the ball lands, many of the red specks will have outpaced the initial 'ball', and many of the blue specks will have lagged behind, so obviously the time between the first red speck hitting the ground and the last blue speck must be greater than if both colours traveled at exactly the same speed. Change the sign of the dispersion coefficient, and the same analogy still works, but with the red specks lagging behind the ball and the blue specks jumping ahead of it.

The only way to actually end up with a shorter pulse would be if you were to send the slower wavelengths into the material first, followed by the faster wavelengths. Then, they would superimpose due to dispersion and come out the other end closer together. Since our pulses are of uniform wavelength "content" throughout their timespan, pulse-lengthening is always required.