Fraction of stimulated photons that escape from a laser cavity

[Azelketh]

How to calculate the fraction of stimulated photons that escape from a laser cavity
with
alpha=0.1 cm^-1
length of cavity=1mm
refractive index of laser cavity=3.2
the photons are assumed to be escaping into air so approx refractive index is 1.

Im assuming that alpha is some sort of stimulation coefficient, as its units indicate a quantity per length, for each photon traveling a distance D number of stimulated photons emitted =Dalpha
(im forced to assume this as I am not given any information on what alpha is supposed to represent, just a quantity and units)

I know that only photons incident upon the laser cavity / air boundary at angles less than the critical angle will escape, but i have no idea how to calculate how many stimluated photons would be incident within that angle range.
Can anyone point me in the correct direction?

 

[Redbelly98]

It looks like alpha is an absorption coefficient. If the photons are never absorbed, then 100% of them will escape from the cavity.

So, a photon will either be absorbed, or it will escape from the cavity--eventually. Part of this problem is in calculating the probability that a photon is absorbed when it traverses the length of the cavity. And also, calculating the probability that it escapes (is transmitted) upon reflection at the end face of the cavity. Since no reflectivity is given, assume an uncoated surface. Also, the angle of incidence will be close to normal.

 

[Azelketh]

Cheers, that put me on the track to a solution.
I assumed that one of the ends of the laser cavity was coated with a mirror allowing all photons incident upon it to be perfectly reflected back allowing them to pass out the other end if within the correct angular range.

I assumed that the angular range of photons that can pass through the end of the laser cavity is 4 theta( where theta is critical angle, initiall 2 theta, but accounting for reflected photons gave me an additional factor of 2).

Then the average distance traveled from any point in the cavity would be 2L, as the sum of distances for relflected and non reflected photons would be 2 assuming all reflected photons emitted in the acceptable angular range would still be in that range at the other end of the laser cavity.

This gave me 1-2L*alpha as the fraction of photons that were absorbed ( cheers for the hint about what alpha could be Redbelly98)

Leaving the total fraction of photons that leave the cavity at

4*theta*(1-2L*alpha) / 2pi

=2*theta*(1-2L*alpha) / pi

which for theta=0.318 radians
fraction of photons=0.202
which seems reasonable to me.(hopefully)

 

[Redbelly98]

Cheers, that put me on the track to a solution.
I assumed that one of the ends of the laser cavity was coated with a mirror allowing all photons incident upon it to be perfectly reflected back allowing them to pass out the other end if within the correct angular range.

Sounds reasonable.

I assumed that the angular range of photons that can pass through the end of the laser cavity is 4 theta( where theta is critical angle, initiall 2 theta, but accounting for reflected photons gave me an additional factor of 2).

I don't think the angular range matters here. We are told the photon was created via stimulated emission, so it should be within the angular range supported by the critical angle and/or any cavity mode or geometry constraints. (Do your class lecture notes or textbook indicated otherwise?)

Then the average distance traveled from any point in the cavity would be 2L, as the sum of distances for relflected and non reflected photons would be 2 assuming all reflected photons emitted in the acceptable angular range would still be in that range at the other end of the laser cavity.

This gave me 1-2L*alpha as the fraction of photons that were absorbed ( cheers for the hint about what alpha could be Redbelly98)

You're welcome. Though it's actually 2Lα that get absorbed, in a single round-trip of the cavity. Perhaps that is what you meant to say?

Leaving the total fraction of photons that leave the cavity at

4*theta*(1-2L*alpha) / 2pi

=2*theta*(1-2L*alpha) / pi

which for theta=0.318 radians
fraction of photons=0.202
which seems reasonable to me.(hopefully)

Well, you haven't accounted for the reflectivity of the uncoated end. Some fraction of the photons are reflected back into the cavity, where they might be absorbed -- or they might escape the cavity the next time they encounter the uncoated face.

 

[Azelketh]

where i said (1-2L*alpha) i thought that 2L*alpha was the fraction of photons that were absorbed, making 1-2L*alpha the fraction that were unaffected. is this correct? or is 2L*alpha the actual number of photons absorbed, which given the small magnitude of the number seems odd to me to have 2*10^-5 photons absorbed, as that would make photon absorption negligible and irrelevant surely?

forgot the reflection part, assuming that most of the light has an angle of incidence close to the normal then the relection ratio =(n1-n2) / (n1 +n2)^2

as n1=3.2
n2=1

R=(2.2/4.2)^2 = 121/441

As R is the fraction of photons that are relected, multiplying the fraction that i obtained previously by (1-R) should give the approx number of photons that pass through.

0.202*(1-121/441)= 0.1465...

so approx 14.7% of total photons pass through initally.

Although I am not sure whether the photons which are reflected should be considered as surely once they have been refleced from the far end of the laser cavity then (1-121/441) of them will then pass through and so on. So if hypothetically given infinite time then all photons within the angular range to do so will pass through?

 

[Redbelly98]

where i said (1-2L*alpha) i thought that 2L*alpha was the fraction of photons that were absorbed, making 1-2L*alpha the fraction that were unaffected. is this correct?

Yes, correct. But you said it was the fraction absorbed earlier.

... or is 2L*alpha the actual number of photons absorbed, which given the small magnitude of the number seems odd to me to have 2*10^-5 photons absorbed, as that would make photon absorption negligible and irrelevant surely?

Your reasoning is correct, 2L*alpha is the fraction that is absorbed in one round trip of the cavity. (BTW, that's an approximation, but it's a valid one in this problem.)

forgot the reflection part, assuming that most of the light has an angle of incidence close to the normal then the relection ratio =(n1-n2) / (n1 +n2)^2

as n1=3.2
n2=1

R=(2.2/4.2)^2 = 121/441

Yes, I agree.

As R is the fraction of photons that are relected, multiplying the fraction that i obtained previously by (1-R) should give the approx number of photons that pass through.

0.202*(1-121/441)= 0.1465...

so approx 14.7% of total photons pass through initally.

Because I disagreed with the fraction you obtained previously, I would also disagree with this or anything based on that previous result. If there is something in your lecture notes or textbook that indicates the angle must be accounted for as you did, that would be another matter.

Although I am not sure whether the photons which are reflected should be considered as surely once they have been refleced from the far end of the laser cavity then (1-121/441) of them will then pass through and so on. So if hypothetically given infinite time then all photons within the angular range to do so will pass through?

No, some of them will get absorbed. In fact, a fraction of 2*L*alpha get absorbed in between consecutive reflections.

So, what we have figured out (and agree on, I think) so far is that, in one round trip of the cavity:

• a fraction (1-121/441) of the photons leave the cavity
• a fraction 2*L*alpha are absorbed.

 

[Azelketh]

Ah, I hadn't cosidered that as the photons are emitted by stimulated emission that they would all be in the angular range requiered (can assume so for this simplified problem).
So that my initial problem's answer is simply
(1-121/441) * (1-2L*alpha)
=(320/441)*( 1-2*10^-5)
=0.7256...
so approx 73% of the photons escape the laser cavity.

Kudos for your aid Redbelly98.

 

[Redbelly98]

Looking better. But L*alpha is not 10^-5.

 

[Azelketh]

whoops, stupid calculation error there,
alpha=0.1 cm^-1 = 10 m^-1
length of cavity=1mm = 10^-3 m
L*alpha=10^-2 and is unitless

(1-121/441) * (1-2L*alpha)
=(320/441)*( 1-2*10^-2)
=(320/441)*(49/50)
=32/45
=0.71111111...
so approx 71% pass through.
heh finally...