# mode-locked laser Repetition Rate

I don't understand how we can produce a laser system with different (lower) repetition rate than the resonant frequency of the cavity?

In other words, when we have different resonating modes, then isn't it that the separation frequency is always c/2L, where L is the distance between the two mirrors (λ = 2L).

So, in other words, let's say if we have a wavelength of λ=1550 nm, then f=c/λ then the frequency of 1550 nm light to be f=193 THz

So, in other words, the separation frequency of the modes will be f=193 THz. Then how can we generate pulses at a repetition rate equal to the bit rate B?

The bit rate (B) is real applications could be way lower than the separation frequency. Imagine that our data rate is only 5 Gbps, then how are we going to produce pulses at that (relatively slower) rate, knowing that inside the cavity, the repetition rate (in this example) is very high @ f=193 THz? (since this is what the resonant frequency is).

How can cavity produces repetition pulses way slower than the resonant frequency?

The repetition rate of the laser ($f_0=c/2L=$ 'bit rate') is the frequency at which pulses circulate in and escape from the cavity. It is also the frequency of the lowest resonant mode, and the frequency separation between modes. This is not the same thing as the frequency of the EM field ($f=c/\lambda$), which is a very large multiple of $f_0$.

The laser is not like an organ pipe oscillating in its lowest resonant mode ($f_0=c/2L$) or a low harmonic ($2f_0, 3f_0,$ etc). It is oscillating in an extremely high harmonic. In your case the lowest mode is $f_0=5$ GHz and the output frequency is $193$ THz which is $38,600 \times f_0$.

Any mode such that $m\lambda=2L$ can resonate. Which modes do resonate depends on which modes are amplified. In this case only optical modes far above the lowest resonant mode $f_0$ are amplified. For mode-locked lasers a broad range of modes are amplified, each being a multiple of $f_0$.

To be technically correct, the $c$ used in $f_0=c/2L$ is the group velocity, the speed of the envelope of the pulse, not the phase velocity of the carrier wave. The effect of this is that there is a round-trip phase slip from one pulse to the next between the carrier and the envelope. Because of this the frequencies in the mode-locked spectrum are offset from exact multiples of $f_0$ by a carrier-envelope phase slip frequency $f_1$ : $$f=mf_0+f_1$$ $L$ is also the optical length of the cavity, taking into account the refractive index of materials in the cavity.

• But isn't it that in order to make a laser, the length (L) has to be some multiple of the wavelength (λ)? So, this indicates that there has to be a relation between 'bit rate' and the frequency. But in a real application, they may use a 1550nm laser, but have any arbitrary bit rate such as 5Gbps, or 4Gbps, or 3Gbps..... How? – Rudy01 Feb 10 '18 at 19:20
• My answer explains. The wavelength of laser light is a multiple of the cavity length ... a very very large multiple. Bit rate/repetition rate is something different. To change the repetition rate you change the length of the cavity. – sammy gerbil Feb 10 '18 at 19:23
• Two organ pipes with different lengths can produce the same frequency if they are resonating at different harmonics of the fundamental frequency. The spacing between modes (=bit rate) will be different because the lengths are different. – sammy gerbil Feb 10 '18 at 19:31
• I see what you are saying, but if you still want to use a 1550 nm laser, then you have to have a laser that resonates at a multiple of that frequency.... If I change the length of the cavity to any arbitrary length, then I cannot make the claim that I am still using a 1550 nm laser? – Rudy01 Feb 10 '18 at 19:37
• In all your answers, I see a relationship between 'bit rate' and 'fundamental frequency'..... How can we generate a 'bit rate' that has no relation to the fundamental frequency...... There is NO multiple integer between frequency of 193 THz and 5GHz ....??? – Rudy01 Feb 10 '18 at 19:47