Range of wavelengths from a laser pulse?

  • Thread starter hb1547
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  • #1
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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?
 

Answers and Replies

  • #2
fzero
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You need to use some calculus to relate the variations. If

[tex]f = \frac{c}{\lambda},[/tex]

then

[tex]\delta f = - \frac{c}{\lambda^2} \delta \lambda.[/tex]

You can deal with the minus sign by computing absolute values.
 
  • #3
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Hmm, so I gave that a try:

[tex]\Delta \omega \Delta t \geq \frac{1}{2}[/tex]

[tex]\delta (\omega )= \delta ( 2 \pi f)[/tex]

[tex]\delta \omega = 2 \pi \frac{-c}{\lambda^2} \delta \lambda[/tex]

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?
 
  • #4
fzero
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I found

( 600 10-9 m)2/(4 pi (3 108 m/s) (10-15 s) ) = 95nm
 
  • #5
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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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