# Find increase in pulse duration for pulse through fused silica

1. Apr 24, 2014

### unseenmisfit

1. The problem statement, all variables and given/known data
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)

2. Relevant equations

3. 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?

Last edited: Apr 24, 2014
2. Apr 27, 2014

### 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-ωc:

Eout = Ein exp[-jø(ω-ωc)]. ωc 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-ωc). Thus the units of ø2 are T2 since ø2 is the coefficient of the third-order variable (ωc - ω)2.

You obviously need this Δtin → Δtout 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.

3. Apr 30, 2014

### 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?

4. Apr 30, 2014

### 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 Δtout 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 Δtout = Δtin 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.

5. Apr 30, 2014

### unseenmisfit

This is what he sent out as the solution... but he wrote 362 instead of 361 so I dont know if he messed up his calculations

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

6. Apr 30, 2014

### rude man

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:
Δtout = √{Δtin4 + 16(ln2)2ø22}/Δtin with ø2(20mm) = 2ø2(10mm). That's the total output pulse width so subtract the 10 fs input duration to get elongation.