# Photon lifetime in an optical cavity

Hi I have the following problem and data for an optical cavity.

Length $$\ L=100 \mu m\$$

Reflectivity $$\R=0.33\$$

Refraction index $$\n=3.6\$$

Inner losses $$\p_i=20 cm^{-1}\$$

Now I have to determine the photon lifetime $$\ \tau_p\$$.

My attempt was to consider the group velocity and then divide it by the length

$$\frac{c}{nL}=8.333 \times10^{11} s^{-1}$$

Taking the inverse of that I get:

$$1.2 ps$$

My question now is how to I relate this time with the actual photon lifetime, and what role do the reflectivity and inner losses play in my expression. Any help is appreciated.

• Perhaps a question more suited to Physics.SE – AJN Nov 18 '20 at 17:58
• Probably won't move it seeing it has an answer and moving erases answers now. – Voltage Spike Nov 18 '20 at 18:56

You've calculated the time for a photon to take one round trip around the cavity. But you haven't considered any of the loss mechanisms, so you haven't determined how many round trips it can make, on average before it is absorbed or emitted out of one of the facets.

You want to use the equation

$$\frac{1}{\tau_p}=v_g \alpha + \frac{v_g}{L}\log\frac{1}{\sqrt{R_1 R_2}}$$

where $$\\tau_p\$$ is the photon lifetime, $$\v_g\$$ is the propagation velocity in the cavity, $$\\alpha\$$ is the distributed loss, $$\L\$$ is the cavity length, and $$\R_1\$$ and $$\R_2\$$ are the mirror reflectivities.

Note that the term $$\\frac{1}{L}\log\frac{1}{\sqrt{R_1 R_2}}\$$ is often called the "distributed mirror loss". That is, it's the loss you would need to add to the material and waveguide loss ($$\\alpha\$$) to give the same effect over one round trip, as the two discrete mirrors.

• Hey! With your expression I get 2.5 ps while, apparently, the actual answer is 0.9 ps. Was I supposed to make R1=R2=R? – Granger Obliviate Nov 18 '20 at 21:07
• @GrangerObliviate, yes, $R_1$ and $R_2$ are the reflectivities of the two mirrors. If you were using a different symbol for that in your problem, you'll have to translate. – The Photon Nov 18 '20 at 21:55
• Is this the average lifespan? If so, how does one calculate the standard deviation? – tuskiomi Nov 19 '20 at 5:13
• @tuskiomi It's an exponential decay (more or less---we're making the approximation that the mirror loss happens throughout the cavity instead of just at the ends) so the standard deviation is equal to the mean. – The Photon Nov 19 '20 at 5:26
• @tuskiomi, from Wikipedia: "the exponential distribution is the probability distribution of the time between events in a Poisson point process". The number of photon absorbtion/emission events is Poisson. The lifetime of an individual photon is exponential. That does not mean that any photon has a negative lifetime (I think you are mis-applying a rule for normal distributions). – The Photon Nov 19 '20 at 16:17