How Can I Calculate the Number of Photons in a Laser Cavity?

Vegeta2019
Messages
2
Reaction score
0

Homework Statement



The laser cavity is formed by two mirrors separated by 15 cm. One of the mirrors has an ultra-high reflectivity and the output mirror has the much lower reflectivity of 99.5 %. How many photons are there in the cavity?[/B]

The power of the laser is 1mW and the wavelength is 600nm

Homework Equations


E= hc/λ and N=P/E Hence N = Pλ/hc

Power out = Power in x (1 - reflectivity) ... Not sure about this formula

The Attempt at a Solution


I have different views to what is happening as I think the question is vague. So I only have worked out certain elements.

At full power N = Pλ/hc ... N=3.01x1015

We could work out the time inside the cavity t = 30x10-2 /3x108 which gives 1 nano second... Not sure if this is relevant to the question.

Using the formula Power out = Power in x (1 - reflectivity) gives a power of 1x10-3 x(1 -0.05)
=9.5x10-4

Using this new power N = Pλ/hc... N= 2.86x1015 photons/s

Would I then do 3.01x1015 - 2.86x1015 = 1.6x1014 ??

Not sure if this is correct. Is the time element I tried supposed to be used?
 
Physics news on Phys.org
I would approach the problem this way: start by calculating the number of photons escaping the cavity per unit time, then figure out how many photons per unit time need to be hitting the semi-reflecting mirror for the calculated output, then figure out how many photons must be inside the cavity to get this hit rate on the mirror.
 
So If I work our the photons escaping the cavity per uni time this would look like

Photons escaping = 3.01x1015 / 1x10-9 =3.01x1024 since the time the spend in the cavity is 1 nano second??
From my attempt the number of photons needed to hit the semi reflecting mirror is N= 2.86x10^15 photons/s due to the 0.5% loss?

So it would be 3.01x1024 - 2.86x1015 = 3x1024

Still not sure if I am doing this correct.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top