Lasers and Fabry Perots - Homework 02 - Lasers and Optomechanics

This is the second homework assignment for Lasers and Optomechanics at Syracuse University.

It is due Monday, February 16, 2026

You will need to complete the questions in this jupyter notebook and submit it via gitlab

Readings

Chapter 1 of Lasers by Seigman: Free eBook

1Atomic Rate Equations¶

Problem 1 from Chapter 1.5 of Lasers by Siegman

Screenshot 2026-01-04 at 12.54.55 AM.png

1.1Atomic Rate Equation Solution¶

From Siegman Chapter 1.5:

\begin{align} \dfrac{d N_2}{dt} \approx R_p - \gamma_{21} N_2\\ \dfrac{d N_1}{dt} = \gamma_{21} N_2 - \gamma_{10} N_1 \end{align}

(1)

where $N_1$ and $N_2$ are the number of atoms in the excited $E_1$ and $E_2$ energy levels,
$\gamma_{ij}$ is the relaxation rate from state $E_i$ to state $E_j$ , and
$R_p$ is a combination of the pumping rate up to $E_3$ and fast relaxation down to $E_2$ .

Now, we add the relaxation $\gamma_{20}$ from $E_2$ to $E_0$ .
This alters the first equation above to

\begin{align} \dfrac{d N_2}{dt} \approx R_p - (\gamma_{21} + \gamma_{20}) N_2 \end{align}

(2)

The $\dfrac{d N_1}{dt}$ is not altered.

In the steady state, $\dfrac{dN_i}{dt} = 0$ . This implies

\begin{align} 0 &= R_p - (\gamma_{21} + \gamma_{20}) N_2\\ N_2 &= \dfrac{R_p}{\gamma_{21} + \gamma_{20}}\\ ~\\ N_1 &= \dfrac{\gamma_{21}}{\gamma_{10}} N_2 \end{align}

(3)

Population inversion requires $N_2 - N_1 > 0$ . Subbing in the above equations

\begin{align} N_2 - N_1 > 0\\ N_2 \left(1 - \dfrac{\gamma_{21}}{\gamma_{10}} \right) > 0\\ \dfrac{R_p}{\gamma_{21} + \gamma_{20}} \left(1 - \dfrac{\gamma_{21}}{\gamma_{10}} \right) > 0\\ \end{align}

(4)

The population inversion requirement can only be satisfied if $\gamma_{10} > \gamma_{21}$ .
This is the exact same condition that was derived in Chapter 1.5 of Siegman.
The new relaxation directly from $E_2$ down to $E_0$ has no effect on the population inversion condition.

This loosely makes sense: while it is true that $N_2$ is lower in the steady state when we include $\gamma_{20}$ , $N_1$ is decreased by the exact same ratio. So this new relaxation path reduces the efficiency of our laser, but does not impact it’s ability to invert it’s lasing population levels.

2Pulsed Laser Power¶

Pulsed Laser Intensity¶

2.1Pulsed Laser Power + Pulsed Laser Intensity Solution¶

Parameters: Density $\rho = 2 \times 10^{19} \mathrm{\dfrac{ions}{cm^3}}$
Pulse Length $\tau = 50~\mathrm{ns}$
Gain Medium Rod Length $L_m = 7.5~\mathrm{cm}$
Gain Medium Rod Radius $R_m = 0.5~\mathrm{cm}$
Ruby Laser Wavelength $\lambda_\mathrm{ruby} = 694.3~\mathrm{nm}$

Gain Medium Volume $V = \pi L R^2$ ,
which will have $N = \rho V$ total ions in the excited state when the Q-switch is blocked.

If all the ions discharge a photon with energy $E_\mathrm{ruby} = h \nu_\mathrm{ruby}$ in a single pulse,
the total energy released is

E_\mathrm{total} = N h \nu_\mathrm{ruby} = N \dfrac{h c}{\lambda_\mathrm{ruby}} = \rho V \dfrac{h c}{\lambda_\mathrm{ruby}} = \rho \pi L R^2 \dfrac{h c}{\lambda_\mathrm{ruby}}

(5)

The peak power in that pulse, assuming a flat-top pulse with equal power over the entire $50~\mathrm{ns}$ , will be

P_\mathrm{peak,flat-top} = \dfrac{E_\mathrm{total}}{t_\mathrm{pulse}} = \rho \pi L R^2 \dfrac{h c}{\lambda_\mathrm{ruby} t_\mathrm{pulse}}

(6)

The peak power in the pulse for a Gaussian pulse profile will be a factor of two higher:

P_\mathrm{peak,gaussian} = \dfrac{2 E_\mathrm{total}}{t_\mathrm{pulse}} = 2\rho \pi L R^2 \dfrac{h c}{\lambda_\mathrm{ruby} t_\mathrm{pulse}}

(7)

Intensity¶

Laser beam radius $w = 0.5~\mathrm{mm}$

For a beam radius $w$ , the peak intensity at the center of the beam will be

I_\mathrm{peak} = \dfrac{2 P}{\pi w^2}~\mathrm{\dfrac{W}{m^2}}

(8)

The peak optical field strength at that spot will be

|E|_\mathrm{peak} = \sqrt{\dfrac{2 I}{c \epsilon_0}}

(9)

which comes from our intensity relationship in Lecture 2:

I = \dfrac{1}{2} c \epsilon_0 |E|^2

(10)

import numpy as np
import scipy.constants as scc

rho = 2e19 # ions/cm^3
lambda_ruby = 694.3e-9 # m
gain_medium_length = 7.5 # cm
gain_medium_radius = 0.5 # cm
t_pulse = 50e-9 # seconds
w_beam_radius = 0.5e-3 # m

volume = np.pi * gain_medium_length * gain_medium_radius**2
NN = rho * volume

total_energy = NN * scc.h * scc.c / lambda_ruby
peak_power_flat = total_energy / t_pulse
peak_power_gaussian = 2 * total_energy / t_pulse
peak_intensity_gaussian = 2 * peak_power_gaussian / (np.pi * w_beam_radius**2)
optical_field_strength_gaussian = np.sqrt(2 * peak_intensity_gaussian / (scc.c * scc.epsilon_0))


print(f"Total Energy          = {total_energy:.1f} J")
print(f"Peak Power (Flat Top) = {peak_power_flat*1e-9:.2f} GW")
print(f"Peak Power (Gaussian) = {peak_power_gaussian*1e-9:.2f} GW")
print()
print(f"Peak Intensity (Gaussian)   = {peak_intensity_gaussian*1e-9:.2f} GW/m^2")
print(f"Peak Optical Field Strength = {optical_field_strength_gaussian*1e-9:.1f} GV/m")

Total Energy          = 33.7 J
Peak Power (Flat Top) = 0.67 GW
Peak Power (Gaussian) = 1.35 GW

Peak Intensity (Gaussian)   = 3433292.57 GW/m^2
Peak Optical Field Strength = 1.6 GV/m

3Geometric Series Fabry-Perot¶

3.1Part A:¶

Rederive the Fabry-Perot intracavity electric field $E_\mathrm{cav}$ using the fact that the infinite geometric series

\begin{align} \sum_{n=0}^\infty x^n = 1 + x + x^2 + \cdots x^n + \cdots = \dfrac{1}{1 - x} \qquad \text{iff} |x| < 1 \end{align}

(11)

Hint 1: Set up some contributing fields $E_n$ for $n$ round-trips.

3.2Part B:¶

Draw a plot of the first couple of electric fields $E_n$ , as well as the total phasor $E_\mathrm{cav}$ ,

while on resonance,
while just off resonance, $\phi_\mathrm{rt} \neq 0$ , but $\phi_\mathrm{rt} \ll 1$ .

3.3Part C:¶

The previous parts we’ve assumed there is zero delay in the propogation time: i.e that the fields in the cavity are in steady state. Now let’s relax this assumption.

What will be the response of the Fabry-Perot intracavity field $E_\mathrm{cav}$ to a step input $E_\mathrm{in}$ ?

For simplicity, assume that the input laser is exactly on resonance, such that $e^{i 2 k L} = 1$

What is the round-trip time delay time $\tau_{rt}$ of the cavity?
How much time $t$ must elapse for the $n$ th term of $E_n$ to start contributing? Write an expression for $n$ in terms of $t$ and $\tau_{rt}$ .
Using a partial geometric series, what is the buildup for the cavity $E_\mathrm{cav}(n)$ after $n$ terms are summed together?
Using the model $1 - \exp(-t / \tau_\mathrm{storage})$ , calculate the cavity storage time $\tau_\mathrm{storage}$ .
Compare your result to the cavity pole $\nu_p$ .

3.4Geometric Series FP Solution¶

Part A:¶

As stated in class, we want to have the sum from $n \in [0, \infty)$ of

E_n = t_1 (r_1 r_2 e^{2 i k L})^n E_in

(12)

Let $x = r_1 r_2 e^{2 i k L}$ , and multiply the entire series by $a = t_1 E_in$ . Because $r_1 r_2 < 1$ , we can apply the geometric series to get our known solution:

E_\mathrm{cav} = \dfrac{t_1}{1 - r_1 r_2 e^{2 i k L}} E_\mathrm{in}

(13)

Part B:¶

On resonance:

Screenshot 2026-02-22 at 11.14.11 AM.png

Slightly off resonance:

Screenshot 2026-02-22 at 11.14.24 AM.png

Part C:¶

Exactly on resonance, we have

E_\mathrm{cav} = \dfrac{t_1}{1 - r_1 r_2} E_\mathrm{in}

(14)

The round-trip delay time $\tau_{rt} = \dfrac{1}{\mathrm{FSR}} = \dfrac{2 L}{c}$
$E_n$ represents the intracavity field contribution from the $n$ -th round-trip, so will start contributing at time $t = n \tau_{rt}$ .
Solving for $n$ gives $n = \dfrac{t}{\tau_{rt}}$ .
The partial geometric series is
$a \sum_{i=0}^{n-1} x^i = a \dfrac{1 - x^{n} }{1 - x}$
(15)
using our definitions of $x$ and $a$ from above, we get
$E_\mathrm{cav}(n) = t_1 E_\mathrm{in} \sum_{i=0}^{n-1} (r_1 r_2 e^{2 i k L})^i = t_1 E_\mathrm{in} \dfrac{1 - (r_1 r_2 e^{2 i k L})^n}{1 - r_1 r_2 e^{2 i k L}}$
(16)
A model like $1 - \exp\left(-\dfrac{t}{\tau_\mathrm{storage}}\right)$ represents a exponentially decaying increase to some maximum value, similar in form to our result above in part three. We can make that relationship explicit:

\begin{align} t_1 E_\mathrm{in} \dfrac{1 - (r_1 r_2 e^{2 i k L})^n}{1 - r_1 r_2 e^{2 i k L}} = E_\mathrm{cav}(0) \left(1 - \exp\left(-\dfrac{t}{\tau_\mathrm{storage}}\right) \right) \end{align}

(17)

Recalling that the maximum steady state on resonance for the intracavity power $E_\mathrm{cav}(0) = \dfrac{t_1 E_\mathrm{in}}{1 - r_1 r_2}$ ,
and we sub in $e^{2 i k L} = 1$ and $n = \dfrac{t}{\tau_{rt}}$ :

\begin{align} t_1 E_\mathrm{in} \dfrac{1 - (r_1 r_2)^n}{1 - r_1 r_2} &= \dfrac{t_1 E_\mathrm{in}}{1 - r_1 r_2} \left(1 - \exp\left(-\dfrac{t}{\tau_\mathrm{storage}}\right) \right)\\ 1 - (r_1 r_2)^n &= 1 - \exp\left(-\dfrac{t}{\tau_\mathrm{storage}}\right)\\ (r_1 r_2)^{\frac{t}{\tau_{rt}}} &= \exp\left(-\dfrac{t}{\tau_\mathrm{storage}}\right)\\ \log\left( (r_1 r_2)^{\frac{t}{\tau_{rt}}} \right) &= -\dfrac{t}{\tau_\mathrm{storage}}\\ \dfrac{t}{\tau_{rt}} \log\left( r_1 r_2 \right) &= -\dfrac{t}{\tau_\mathrm{storage}}\\~\\ \rightarrow \tau_\mathrm{storage} &= \dfrac{ \tau_{rt} }{ \log\left( \dfrac{1}{r_1 r_2} \right) } \end{align}

(18)

From class, the Fabry-Perot cavity pole $\nu_p = \dfrac{1}{2\pi} \dfrac{c}{2 L} \log\left(\dfrac{1}{r_1 r_2}\right)$ .

Since $\mathrm{FSR} = \dfrac{c}{2 L} = \dfrac{1}{\tau_{rt}}$ , our result shows that

\tau_\mathrm{storage} = \dfrac{2 \pi}{\nu_p}

(19)

4Finesse and Loss in a Fabry-Perot¶

A very convenient relationship between total loss in a cavity $\mathcal{L}_\mathrm{total}$ and cavity finesse $\mathcal{F}$ can be calculated to be (in the high-finesse limit $\mathcal{F} \gg 1$ ):

\begin{align} \mathcal{F} = \dfrac{2 \pi}{\mathcal{L}_\mathrm{total}} \end{align}

(20)

Derive this result starting with $\mathcal{F} = \dfrac{\mathrm{FSR}}{\nu_\mathrm{FWHM}}$ .

Hint 1: Total loss includes transmission losses for both mirrors: $\mathcal{L}_\mathrm{total} = T_1 + T_2 + \mathcal{L}_1 + \mathcal{L}_2$ .

Hint 2: Write $r = \sqrt{1 - T - \mathcal{L}}$ .

Hint 3: Use the binomial approximation.

Hint 4: This paper from MIT may be helpful: Loss in long-storage-time optical cavities

4.1Finesse and Loss in FP Solution¶

Because we are in the high finesse limit, the reflectivity of the mirrors $r_i$ should be very close to 1, so $1 - r_1 r_2 \ll 1$ :

\begin{align} \mathcal{F} &= \dfrac{\mathrm{FSR}}{\nu_\mathrm{FWHM}}\\ \mathcal{F} &= \dfrac{\dfrac{c}{2 L}}{\dfrac{c}{L \pi} \arcsin\left( \dfrac{1 - r_1 r_2}{2 \sqrt{r_1 r_2} } \right)}\\ \mathcal{F} &= \dfrac{\pi}{2} \dfrac{1}{\left( \dfrac{1 - r_1 r_2}{2 \sqrt{r_1 r_2} } \right)}\\ \mathcal{F} &= \dfrac{\pi \sqrt{r_1 r_2}}{1 - r_1 r_2}\\ \end{align}

(21)

Focusing on $1 - r_1 r_2$ and applying the binomial approximation:

\begin{align} 1 - r_1 r_2 &= 1 - \sqrt{1 - T_1 - L_1}\sqrt{1 - T_2 - L_2}\\ 1 - r_1 r_2 &\approx 1 - \left( 1 - \dfrac{1}{2} \left(T_1 + L_1\right) \right) \left( 1 - \dfrac{1}{2} \left(T_2 + L_2\right) \right)\\ 1 - r_1 r_2 &\approx 1 - \left( 1 - \dfrac{1}{2} \left( T_1 + T_2 + L_1 + L_2 \right) + \dfrac{1}{4}\left( T_1 T_2 + T_1 L_2 + L_1 T_2 + L_1 L_2 \right)\right) \end{align}

(22)

Ignoring all the very small $T_1 T_2$ product terms on the right-hand side yields the approx we want:

\begin{align} 1 - r_1 r_2 &\approx \dfrac{1}{2} \left( T_1 + T_2 + L_1 + L_2 \right) \end{align}

(23)

We can apply this directly to our finesse approx above, and correctly assume that $\sqrt{r_1 r_2} \approx 1$ :

\begin{align} \mathcal{F} &= \dfrac{\pi \sqrt{r_1 r_2}}{1 - r_1 r_2}\\ \mathcal{F} &= \dfrac{\pi}{\dfrac{1}{2} \left( T_1 + T_2 + L_1 + L_2 \right)}\\ \mathcal{F} &= \dfrac{2\pi}{T_1 + T_2 + L_1 + L_2}\\ \mathcal{F} &= \dfrac{2\pi}{ \mathcal{L}_\mathrm{total} } \end{align}

(24)

5Finesse and Gain in a Fabry-Perot¶

5.1Part A:¶

Assuming that we have a critically-coupled Fabry-Perot cavity,
what is the comparison between the finesse $\mathcal{F}$ and power gain $G_\mathrm{cav}$ ?

5.2Part B:¶

Repeat Part A above for an over-coupled cavity such that $r_2 \approx 1$ .
Does the relationship change?

5.3Finesse and Gain in a FP Solution¶

Recall that for the finesse $\mathcal{F}$ and power gain at resonance $G_\mathrm{cav}(0)$

\begin{align} \mathcal{F} &= \dfrac{\pi \sqrt{r_1 r_2} }{1 - r_1 r_2}\\ G_\mathrm{cav}(0) &= \left(\dfrac{t_1}{1 - r_1 r_2} \right)^2 \end{align}

(25)

Part A Solution¶

Assuming $r_1 = r_2 = r$ and no losses such that $t = \sqrt{1 - r^2}$ , we get

\begin{align} \mathcal{F} &= \dfrac{\pi \sqrt{r^2} }{1 - r^2} = \dfrac{\pi r }{1 - r^2}\\ G_\mathrm{cav}(0) &= \left(\dfrac{\sqrt{1 - r^2}}{1 - r^2} \right)^2 = \dfrac{1}{1 - r^2} \end{align}

(26)

then their ratio becomes

\begin{align} \text{Critically-Coupled:}~\dfrac{\mathcal{F}}{G_\mathrm{cav}(0)} &= \pi r \\ &= \pi \sqrt{1 - T}\\ &\approx \pi \left(1 - \dfrac{1}{2} T \right)\\ &\approx \pi \end{align}

(27)

for small $T$ .

Part B Solution¶

Setting $r_2 = 1$ and $t_1 = \sqrt{1 - r_1^2}$ yields

\begin{align} \mathcal{F} &= \dfrac{\pi \sqrt{r_1} }{1 - r_1}\\ G_\mathrm{cav}(0) &= \left(\dfrac{\sqrt{1 - r_1^2}}{1 - r_1} \right)^2 = \dfrac{1 - r_1^2}{(1 - r_1)^2} = \dfrac{1 + r_1}{1 - r_1} \end{align}

(28)

then their ratio becomes

\begin{align} \text{Over-Coupled:}~\dfrac{\mathcal{F}}{G_\mathrm{cav}(0)} &= \pi \dfrac{ \sqrt{r_1} }{1 + r_1}\\ &= \pi \dfrac{ \sqrt{\sqrt{1 - T}} }{1 + \sqrt{1 - T}}\\ &= \pi \dfrac{ (1 - T)^{1/4} }{1 + \sqrt{1 - T}}\\ &\approx \pi \dfrac{ 1 - \dfrac{1}{4} T }{1 + 1 - \dfrac{1}{2} T}\\ &\approx \dfrac{\pi}{2} \end{align}

(29)

So there is a factor of 2 difference between the critically-coupled and over-coupled cases.

6Reflection Phase Angle vs Frequency¶

6.1Plots¶

Make Bode plots of the reflection phase vs round-trip phase where $\phi$ serves as the x-axis.
Hint: This is similar to our expressions $\theta(\phi)$ , where $\theta$ is the reflection phase, and $\phi$ is the round-trip phase.

6.2Reflection Phase Angle vs Frequency Solution¶

The Fabry-Perot reflection field $E_\mathrm{refl}$ is

\begin{align} E_\mathrm{refl} = \dfrac{r_1 - r_2 e^{-i \phi}}{1 - r_1 r_2 e^{-i \phi}} \end{align}

(30)

If $R_2 = 100\%$ , then $r_2 = 1$ and we can simplify the reflection expression:

\begin{align} E_\mathrm{refl} = \dfrac{r_1 - e^{-i \phi}}{1 - r_1 e^{-i \phi}} \end{align}

(31)

Explicitly calculating the magnitude gives

\begin{align} \left| E_\mathrm{refl} \right| &= \left| \dfrac{r_1 - e^{-i \phi}}{1 - r_1 e^{-i \phi}} \right|\\ &= \sqrt{ \dfrac{r_1 - e^{-i \phi}}{1 - r_1 e^{-i \phi}} \dfrac{r_1 - e^{i \phi}}{1 - r_1 e^{i \phi}} }\\ &= \sqrt{ \dfrac{r_1^2 + 1 - r_1 (e^{i \phi} + e^{-i \phi}) }{1 + r_1^2 - r_1 (e^{i \phi} + e^{-i \phi})} }\\ &= 1 \end{align}

(32)

But the phase is not trivial. Finding the real and imaginary parts of $E_\mathrm{refl}$ :

\begin{align} \mathrm{Re}[E_\mathrm{refl}] &= \dfrac{1}{2}( E_\mathrm{refl} + E_\mathrm{refl}^* )\\ \mathrm{Re}[E_\mathrm{refl}] &= \dfrac{2 r_1 - (1 + r_1^2)\cos(\phi)}{1 + r_1^2 - 2 r_1 \cos(\phi)}\\ ~\\ \mathrm{Im}[E_\mathrm{refl}] &= \dfrac{1}{2i}( E_\mathrm{refl} - E_\mathrm{refl}^* )\\ \mathrm{Im}[E_\mathrm{refl}] &= \dfrac{ (r_1^2 - 1) \sin(\phi)}{1 + r_1^2 - 2 r_1 \cos(\phi)}\\ \end{align}

(33)

So the argument of $E_\mathrm{refl}$ must be

\begin{align} \angle E_\mathrm{refl} = \dfrac{1}{1 + r_1^2 - 2 r_1 \cos(\phi)} \arctan2\left[(r_1^2 - 1) \sin(\phi) , 2 r_1 - (1 + r_1^2)\cos(\phi) \right] \end{align}

(34)

This result is plotted below.

def erefl(phi, trans1, trans2, ein=1):
    """Reflected electric field from a Fabry-Perot optical cavity Erefl
    Inputs:
    -------
    phi: float
        round-trip phase of the optical cavity
    trans1: float
        power transmission of the input mirror
    trans2: float
        power transmission of the end mirror
    ein: float
        input electric field, default is ein = 1 sqrt(W)

    Output:
    -------
    erefl: complex float
        complex reflection field from the Fabry Perot
    """
    r1 = np.sqrt(1 - trans1)
    r2 = np.sqrt(1 - trans2)
    erefl = (r1 - r2 * np.exp(-1j * phi))/(1 - r1 * r2 * np.exp(-1j * phi))
    return erefl

phis = np.linspace(-3*np.pi, 3*np.pi, 10000)
trans1 = 0.1 # input power transmission
trans2 = 0 # perfectly reflecting

erefls = erefl(phis, trans1, trans2)

import matplotlib.pyplot as plt

fig, (ax1, ax2) = plt.subplots(2, sharex=True)

ax1.plot(180/np.pi * phis, np.abs(erefls))
ax2.plot(180/np.pi * phis, np.angle(erefls, deg=True))

ax1.set_ylim(0, 1.1)
ax2.set_yticks([-180, -90, 0, 90, 180])
ax1.set_xticks(180*np.arange(-3,4))

ax1.grid()
ax2.grid()

ax1.set_title("Fabry Perot Reflected E-field for Gires-Tournois Interferometer")
ax1.set_ylabel(r"Magnitude $|E_\mathrm{refl}|$")
ax2.set_ylabel(r"Phase $\angle E_\mathrm{refl}$")
ax2.set_xlabel(r"Round-trip phase")