2
$\begingroup$

Before asking my question, I will illustrate how our lecturer treated a Fabry Perot interferometer, because my question is linked to this topic.

Of all the different models to describe laser dynamics, our Laserphysics-lecturer employed a simple semiclassical picture, which treated the electromagnetic field classical, and just in 1 dimension: $E(t, z)$.

The first thing we did then was to look at a Fabry Perot interferometer: If a Beam is propagated into a cavity, consisting of two plane mirrors, and you assume, that at at the two mirros, the beam is just changing its direction of propagation, and the amplitude diminishes by a factor $R_1$ or $R_2$ because of the finite reflectivity of the mirrors.

If the two mirrors are spaced at distances L, and at time $t = 0$ the Field inside the cavity is given by $E_0(t, z)=e^{ikz}$ then after going one roundtrip, the field at every point in the cavity is the same as it was before, times a factor $R_1 R_2 e^{-i2Lk}$

$$ E_1(t, z) = E_0(t,z) R_1 R_2 e^{-i2Lk} $$

If you further assume, that after each roundtrip the external source will still continue to shed light into the cavity, THEN you can assume the whole field in the cavity to be a superposition of all consecutive $E_i$, so the final field (after infinite time) would be:

$$ E(t, z) = \sum_0^\infty (R_1 R_2 e^{-i2Lk})^n E_0(t, z) $$

A geometric series then provides that $E(t, z) = \frac{1}{1-(R_1 R_2 e^{-i2Lk})}E_0(t,z)$

So much for the fabry perot interferometer. What I understand from that is that the intensity inside the cavity will be high if the $2Lk$ is near a multiple of $2\pi$, and that's called resonance.

Now comes the part with my question: From here on one argues, that modes with a frequency $n \frac{c}{2L}$, ($n$ being an integer) are "stable", and other modes are "less stable", and because of that, if I turn on a laser (I put an active material into the cavity), WITHOUT an additional external light source, modes that are off-resonant will decay faster, and modes that are resonant will be amplified more. What I am also told is that "The modes without resonance frequency will destructively interfere".

Why is that? In the previous discussion, to derive the intensity peaks for the resonance frequency $n \frac{c}{2L}$, we have assumed an external source to constantly "bring new light into the cavity". Within a laser, this is not the case, I just have "one" electromagnetic field, going back and forth, following the used equation

$$ E_{n+1}(t, z) = E_{n}(t,z) R_1 R_2 e^{-i2Lk} $$ and being amplified. The amplification over the distance $2L$ has to compensate the losses at the mirrors, but that aside, couldn't a field inside the cavity have any frequency? Why is the phase of the field crucial to whether it's amplified, or decays?

$\endgroup$
1
$\begingroup$

Let's leave the amplifying medium out for the time being, and talk about the Fabry-Perot.

For partially silvered mirrors, the field at the mirrors must be zero. (The details of a dielectric mirror are different, but the result is the same.)

Imagine that the input source is initially off. Then turn it on. If the mirrors merely reflect and transmit, that is, there is no dissipative loss in the mirrors, then the intensity coming out of the interferometer has to be the same as that going in, eventually. (Conservation of energy)

Consider the condition in which we are half way between resonances, and there is a null at the first mirror. Then at the second mirror, there is a peak. In order to zero the field at that mirror, the induced polarization in the second ("exit") mirror generates a field that cancels the incoming field at the mirror. It also cancels the field everywhere else in the cavity. The polarization of the mirror also drives a field that leaves the cavity. It is the same intensity as the input field, and it appears immediately.

At resonance, there is a node at both mirrors. The field at the mirrors is close to zero. It's not exactly zero because of the non-zero transmission. There must be some polarization to drive the field that is transmitted. The reflected wave is in phase, and with each additional pass the field adds to itself, getting larger and larger. During this (very short) time interval, the intensity coming out is less than the intensity going in. Energy is being stored in the cavity. As the intensity inside the cavity builds up, so too does the intensity of the light leaving the cavity. This is because that small polarization in the exit mirror is being driven harder and harder by the field inside the cavity. Initially, there is practically zero exit radiation. Eventually, the rate that energy leaves is equal to the rate of energy entering: steady state is achieved.

Compare the two cases. Off-resonance, radiation leaves the cavity immediately. On resonance, energy is stored. Off-resonance the cavity appears to be very lossy. On resonance, much much less lossy. Attributing the effect to destructive interference is a short-hand, and rather glib, way of expressing all this.

For a laser oscillator, there is no input field. The role of the input field is played by spontaneous emission. The fields produced by spontaneous emission will be of any and all possible wavelengths and phases within the gain band. Radiation from spontaneous emissions that is in resonance will stay in the cavity and grow by stimulated emission and the mechanism I described. Those that are not will immediately leak out after undergoing a small amount of stimulated emission. The power balance is now between the pumping energy and the exiting radiation, but there are other energy paths for the pumping energy. It does not go entirely to the exiting radiation. That is, the conversion efficiency is not 100%.

$\endgroup$
  • $\begingroup$ I hope I can describe my thoughts on your last paragraph right: In order to "grow" by the mechanism your described, the field has to (at least 1 time) interfere with its own reflection. Does that mean, the field that is produced by spontaneous emission "lasts" longer than one cavity round trip? $\endgroup$ – Quantumwhisp Jan 25 '17 at 22:38
  • $\begingroup$ Or is this not about interference at all, just about the reflection properties of a mirror? $\endgroup$ – Quantumwhisp Jan 26 '17 at 18:02
  • $\begingroup$ It's more about interference. I added the stuff about the mirror to elucidate how the reflection leads to interference in the first place. But the interference wouldn't be there without the mirrors, so maybe I should say that it's about both. $\endgroup$ – garyp Jan 26 '17 at 18:14
  • $\begingroup$ Then my question remains: If the field shows interference effects with itself, then the field has to "last" longer than cavity roundtrip? $\endgroup$ – Quantumwhisp Jan 26 '17 at 18:56
  • $\begingroup$ Interference will commence as soon as the field hits the mirror and starts back the opposite direction. But for the full-wave interference that one normally sees pictured, yes. The field has to "last" longer than one round trip. $\endgroup$ – garyp Jan 26 '17 at 19:46
0
$\begingroup$

Firstly why do you think that a laser doesn't have an external source generating new E-fields? A laser must include external excitation to achieve the energy input threshold required for lasing to start. If this is removed, the laser will switch off.

Fields of any frequency can of course exist in the cavity, but Fields of frequencies that do not reinforce positively when they interfere with their reflection will attenuate rather than accumulate over multiple cycles/reflections (i.e. resonate).

$\endgroup$
  • 1
    $\begingroup$ A laser oscillator doesn't need an external field source. It needs an external pump. The initial field can (and usually does) come from spontaneous emission. But an external input field is perfectly valid. There are plenty of laser amplifiers that operate using an external input field. $\endgroup$ – garyp Jan 25 '17 at 14:41
  • $\begingroup$ by "external" I just mean energy needs to be put in, however it is done. I hope that's not misleading. In this question the model has already been specified as mirrors around a cavity and a light source. I'm not sure it's helpful/necessary to bring in things like the spontaneous or stimulated emission from media around the cavity, or absorption and re-emission. $\endgroup$ – JMLCarter Jan 25 '17 at 14:54
  • $\begingroup$ I edited it, I hope it's more clear now. For the Fabry Perot Interferomter, the model is mirrors and an external light source. However, I explicitly asked for the mechanism inside a laser, without any external light source. $\endgroup$ – Quantumwhisp Jan 25 '17 at 16:04
  • $\begingroup$ Whether the light source is physically external or internal has no bearing on the frequency at which a cavity of specific dimensions resonates, in the model you asked about. All that matters is that enough energy is added each cycle to at least overcome reflective losses - there must be a source. $\endgroup$ – JMLCarter Jan 25 '17 at 16:48
  • $\begingroup$ In a real laser the cavity walls are of a material the emits at a particular frequency of light matching the resonance frequency of the cavity. Although there are many types of Laser. $\endgroup$ – JMLCarter Jan 25 '17 at 16:50
0
$\begingroup$

Let me simplify the answer for the sake of future readers who may have the same question.

The destructive interference picture is indeed the most intuitive way of understanding what happens in the cavity without regard to the microphysics (by which I mean the actual microscopic mechanism of the process). To get this, take a plane wave of angular frequency $w$ different from the cavity resonance $w_0$. Then,

\begin{equation} \label{eq:1} \tag 1 E(z,t) = E_0 e^{j(kz-wt)} \end{equation} After a round trip, this becomes,

\begin{equation} \label{eq:2} \tag 2 E(z+2L,t) = E_0 e^{j(k(z+2L)-wt)} \end{equation}

Adding \eqref{eq:1} and \eqref{eq:2}, and simplifying, we get

\begin{equation} \label{eq:3} \tag 3 E(z,t) + E(z+2L) = E_0e^{j(kz-wt)}(1+e^{j2kL}) =2E(z,t)e^{jkL}\cos(kL) \end{equation}

However, using the facts that, $$ k = w/c \\ 2L = \frac{c}{T} \implies L = \frac{c}{2T} $$ where $T$ is the round-trip time, we get, \begin{equation} \label{eq:4} \tag 4 KL = \frac{wT}{2} \end{equation}

With \eqref{eq:4}, we express \eqref{eq:3} as $$ \label{eq:5} \tag{5} E\left(z,t\right)+E\left(z+2L\right) = 2E\left(z,t\right) \exp{\left(j \frac{wT}{2}\right)} \cos{\left(\frac{wT}{2}\right)} $$

Calling the LHS of \eqref{eq:5} $E'(z,t)$ as suggested by the RHS of that eqn., with $w$ and $T$ known constants, we have \begin{equation} \tag{6} E'(z, t) = 2E(z,t) \exp{\left(j \frac{wT}{2}\right)} \cos{\left(\frac{wT}{2}\right)} \end{equation} Suppose then that, $$ w = w_0 - \Delta w \tag{7} \label{eq:7} $$ Where $\Delta w$ is the cavity detuning. Then, $$ E' \left(z, t\right) = 2E\left(z,t\right) \exp{\left( j \frac{w_0 T}{2} \right)} \exp{\left( j \frac{\Delta w T}{2} \right)} \cos{\left( \frac{w_0 T - \Delta w T}{2} \right)} \tag{8} \label{eq:8} $$ But, since by definition $w_0 T = m2 \pi$ for $m = 1,2,3, \dots$, \eqref{eq:8} therefore becomes, $$ E'\left(z, t\right) = {\left(-1\right)}^{m} \cdot 2E\left(z,t\right) \exp{\left(j \frac{\Delta w T}{2} \right)} \cos{\left(\frac{\Delta w T}{2}\right)} \tag{9} \label{eq:9} $$ Here then is the crux of the matter: as long as $\Delta w \approx 0 ,$ meaning $w \approx w_0 ,$ $\Delta w T$ would have any value except $0$ and $m2\pi ,$ $m =1,2,3,\dots .$

Now, since $\left| \cos{\left(\Delta w T / 2\right)}\right| \le 1 ,$ with the equality holding iff $\Delta w T = 0~\text{or}~m2 \pi ,$ it follows simply that $$ \left|E'\left(z, t\right)\right| < \left|E\left(z,t\right)\right| \,.$$ This is the destructive interference of a wave field with a delayed copy of itself. In particular, if$$ \Delta w T = \left(2m+1\right) \pi \,, \qquad \rlap{m = 0,1,2,\dots} $$then we have $E'\left(z, t\right) = 0 ,$ which is complete destructive interference between a wave field and a delayed copy of itself.

Thus, in the general case, we have that off-resonance a wave field can not build up in a cavity due to destructive interference between the wave field and a delayed copy of itself and, in the special case where the round-trip phase accumulated by the delayed copy of the wave field is an odd integer multiple of $\pi ,$ the field in the cavity is zero at all times due to complete destructive interference between the wave field and a delayed copy of itself. Note what this means. Interference is not a destruction of energy, but a redistribution of it.

Accordingly, since a physical wave carries energy in some form, then in the general case described above, some small amount of the energy injected into the cavity via the input wave field is trapped in the cavity while a larger portion of it is reflected out of the cavity, whereas in the special case described above 100% of the energy is reflected out of the cavity and none is trapped in it.

$\endgroup$
  • 1
    $\begingroup$ @Puzzledstudent Wonderful edit! $\endgroup$ – Nat Jun 11 at 22:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.