Range of wavelengths from a laser pulse?

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Homework Help Overview

The problem involves determining the range of wavelengths in a short laser pulse, specifically a 1 femtosecond pulse of laser light that is approximately 0.3 micrometers long, derived from a 600 nm laser source.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the uncertainty principle and manipulate equations relating frequency and wavelength, but expresses confusion over discrepancies with the expected answer. Other participants suggest using calculus to relate frequency and wavelength variations, and question whether the original poster missed a power of ten in their calculations.

Discussion Status

Participants are actively engaging with the problem, exploring different mathematical approaches and clarifying concepts. Some have provided guidance on using calculus to derive relationships, while others are verifying their calculations and discussing potential errors.

Contextual Notes

The discussion includes references to specific equations and constants, as well as the original poster's uncertainty about their calculations and the validity of their approach. There is a noted discrepancy between the calculated and expected results, prompting further investigation.

hb1547
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Homework Statement


"A 1 fs pulse of laser light would be 0.3 um long. What is the range of wavelengths in a 0.3 um long pulse of (approximately) 600nm laser light?"


Homework Equations


(delta omega)(delta t) >= 1/2
c = (lambda)(frequency)

The Attempt at a Solution


I replaced (delta omega) with 2*Pi*(Delta F), which I then replaced with 2*Pi*c/(Delta Lambda).

Then, solving for (Delta Lambda), I got:

Delta lambda = 4*Pi*c*(delta T).

This gives me 3.77*10^(-6) m, which is 3.77 um.

However the book claims the answer is 95 nm... I'm not really getting what I did wrong? Is this completely the wrong approach?
 
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You need to use some calculus to relate the variations. If

f = \frac{c}{\lambda},

then

\delta f = - \frac{c}{\lambda^2} \delta \lambda.

You can deal with the minus sign by computing absolute values.
 
Hmm, so I gave that a try:

\Delta \omega \Delta t \geq \frac{1}{2}

\delta (\omega )= \delta ( 2 \pi f)

\delta \omega = 2 \pi \frac{-c}{\lambda^2} \delta \lambda

Then, if I solve that for d(lambda), and plug everything in, I get about .95 um, which is an order of magnitude higher than what I wanted. Did I miss a power of 10 somewhere, or is that just a coincidence?
 
I found

( 600 10-9 m)2/(4 pi (3 108 m/s) (10-15 s) ) = 95nm
 
hmmm and this time I did get that, I must've plugged it into Mathematica wrong.

Thank you for your help! :]
 

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