Find increase in pulse duration for pulse through fused silica

[unseenmisfit]

Homework Statement

Find increase in pulse duration for 10 fs laser pulse going through 20 mm of fused silica
(\phi^{(2)})=361.626 fs^{2} for 10 mm of fused silica)

Homework Equations

The Attempt at a Solution

I have no idea where to begin with this. Can anyone give me some equations or send me to a page that will help me out?

 

[rude man]

I had to do some research in for this so don't expect full understanding on my part:

The basis for the pulse stretching is the finite value of d²n/dλ² in the material. The method is to Fourier-transform the Gaussian electric field pulse and determine its phase spectrum. This gives the usual input-output relationship in terms of a magnitude (attenuation) and phase, both as functions of frequency ω. The transform is in terms of ω-ω_c:

E_out = E_in exp[-jø(ω-ω_c)]. ω_c is the carrier frequency and ω is the Fourier frequency variable.

For a given medium length there is a relation between the incoming pulse width and the outgoing pulse width which includes the coefficient ø₂ of the third term in a Taylor series expansion of the spectral phase around the carrier frequency ø(ω-ω_c). Thus the units of ø₂ are T² since ø₂ is the coefficient of the third-order variable (ω_c - ω)².

You obviously need this Δt_in → Δt_out formula and I can't give you that without violating forum rules. But you should be able to find it in your textbook or in the Internet as I did.

You're given ø₂ for 10 mm of fused silica and the incoming pulse width of 10 fs, so you can now find the output pulse width after passing thru 20 mm of fused silica.

Hint: ø₂ = k(ω)L, k(ω) = propagation constant and L = length of the material. Should make it easy to modify the output pulse width from the L=10 mm case to the L=20 mm one.

 

[unseenmisfit]

Okay now I see it in my notes, thank you. Just need someone to explain it.
So \frac{∂}{∂ω}\frac{1}{v_{g}} at ω_{0} is k(ω) and is a constant so k(ω)*L*\pi/t_{in}=t_{out}
\frac{361.626 fs^{2}}{10 mm}*\frac{20 mm* \pi }{10 fs}=227.216 fs with the final answer being 227.216fs-10fs=217.216 fs
Is that correct?

 

[rude man]

My hint was off. Sorry. Should have said ø₂ = k₂(ω)L, not k(ω)L.
But the rest of my post I think was right. Your computation of Δt_out might have been based on that error. Sorry again!

Anyway, the hint that was meant was simply that ø₂(20mm) = 2ø₂(10mm).

My answer for L=20mm was 200.777 fs if I did the math right. It also reduces to Δt_out = Δt_in if ø₂ = 0 fs² which is nice!

The theory involves ø₂ which you seemingly neglected. Your formula needs to account for the rate of dispersion with respect to λ (= d²n/dλ²) which is the basis for the pulse stretching phenomenon. That's where ø₂ comes in.

BTW since the problem asks for the increase in pulse duration I guess you should subtract 10 fs from my answer. Could you post the answer when you get it from the prof? Thanks.

 

[unseenmisfit]

This is what he sent out as the solution... but he wrote 362 instead of 361 so I don't know if he messed up his calculations

\Delta T_{GDD} \approx \phi^{(2)} \Delta\omega = 362.626 fs^2 \times \frac {\pi}{10 fs} = 227.45 fs
Increase = 217.45 fs

 

[rude man]

This is what he sent out as the solution... but he wrote 362 instead of 361 so I don't know if he messed up his calculations

\Delta T_{GDD} \approx \phi^{(2)} \Delta\omega = 362.626 fs^2 \times \frac {\pi}{10 fs} = 227.45 fs
Increase = 217.45 fs

Thanks. Does not quite jive with my formula. But when it comes to Gaussian pulses the definition of "pulse width" can vary among folks. In theory the output pulse never vanishes completely. Even if the input pulse has infinitely fast edges, the output gets blurred all the way to infinity. So depending on the cutoff criterion I guess you can get different answers. At least we were off by only about 10%.

My formula was more involved:
Δt_out = √{Δt_in⁴ + 16(ln2)²ø₂²}/Δt_in with ø₂(20mm) = 2ø₂(10mm). That's the total output pulse width so subtract the 10 fs input duration to get elongation.