00:01
So we are given that a pulse of light is created by the superposition of a lot of waves.
00:10
That's just the usual stuff.
00:15
And then that we have a pulse that its wavelength is 600 nanometers and lasts 6 .50 seconds.
00:29
And with this, we want to know the center frequency of this pulse of light.
00:32
So that's quite simple.
00:35
So that's a.
00:41
And the center also, it's f0.
00:44
That's just light velocity over the wavelength.
00:51
So this is three times 10 to the 8 meters per second.
00:58
And the wavelength is 600 times 10 to the 9 meters.
01:04
This will give 5 times 10 to the 14.
01:12
That's the central frequency.
01:15
Okay, so then we want to know how many oscillations.
01:21
The wave completes during the pulse and the pulse is 6 femtose seconds long.
01:29
So we just need to take into account that for b, well, this is a pulse, it's a wave, so it fulfills the fact that the bandwidth times the duration is of the other of 1, okay? and therefore, knowing that the central frequency is this, we know the time period is simply one over the central frequency, which is around 2 times, exactly.
02:16
2 times 10 to the minus 15 seconds.
02:23
It's really fast.
02:26
And the number of cycles in the pulse is simply the duration over this time period.
02:34
So we split the duration in periods.
02:38
And we have delta t over t.
02:42
And this is the 6 times 10 to the minus 15 over 2 times 10 to the minus 15 seconds.
02:49
It's just three cycles.
02:51
Cycles or oscillations.
02:56
Then for c we want to know the range of frequencies...