- #1
StudentonaOdyssey
- 7
- 1
- Homework Statement
- GER:
Zwei bandbreitenlimitierte Pulse mit derselben Mittenfrequenz durchtreten dasselbe transparente optische Bauelement mit einer Dicke von 1 cm und einer GVD von 50 fs2/mm. Der eine Puls ist beim Eintritt 10-mal so lang wie der andere. Welcher der beiden Pulse ist nach dem Durchtritt unmittelbar nach dem Element länger? Um wievielmal länger ist dieser Puls als der andere?
ENG:
Two bandwith-limited pulses with the same middle frequency pass through the same transparent optic component with a thickness of 1cm and a GVD (Group Velocity Dispersion) of 50 fs^2/mm. One pulse is 10-times as long as the other on impact. Which one of the pulses is longer directly after the component? How many times longer is this pulse than the other.
- Relevant Equations
- Bandwidth-Limit
$$\Delta \omega \cdot \Delta t >= C$$
Pulse length after propagation
$$\tau _p = \tau _0 \sqrt{1+\frac{z^2}{D^2}}$$
$$D= \frac{\tau_0^2}{2\cdot GVD}$$
$$\tau _{01} = 10 \tau _{01}$$
If I calculate ##\frac{\tau_{p1}}{\tau_{p1}}## and set z=d=1cm I do not know how to continue from there as I can't solve the equation without knowledge of τ0 for D.
$$\frac{\tau_{p1}}{\tau_{p1}} = \frac{\tau_{02} \cdot 10}{\tau_{02}} \sqrt{\frac{1+\frac{d^2 \cdot 4 \cdot GVD^2}{\tau _{02}^4 \cdot 10^4}}{{1+\frac{d^2 \cdot 4 \cdot GVD^2}{\tau _{02}^4}}}}$$
If I calculate ##\frac{\tau_{p1}}{\tau_{p1}}## and set z=d=1cm I do not know how to continue from there as I can't solve the equation without knowledge of τ0 for D.
$$\frac{\tau_{p1}}{\tau_{p1}} = \frac{\tau_{02} \cdot 10}{\tau_{02}} \sqrt{\frac{1+\frac{d^2 \cdot 4 \cdot GVD^2}{\tau _{02}^4 \cdot 10^4}}{{1+\frac{d^2 \cdot 4 \cdot GVD^2}{\tau _{02}^4}}}}$$
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