The dynamical equations of a laser, based on a four-level atom, are the followin

Question

The dynamical equations of a laser, based on a four-level atom, are the following: dN/dt = R - N(I sigma/h upsilon + 1/tau_2) dI/dt = c sigma NI - I/tau_c a) The laser optical cavity is closed, on one side, bya perfect reflecting mirror and, on the other side, by a partially reflecting mirror. The latter mirror has a small transmission coefficient T. Show that the photon decay rate in the cavity is 1/tau_c c(alpha + T/2L), where alpha is the loss coefficient per unit of length and L the length of the cavity. b) If alpha = 0, show that the maximum intensity at the output of the cavity, when the laser is above threshold, I_out (T) = TI(T), is obtained for T Rightarrow 0. Explain why this result is not physically acceptable. c) We consider now alpha

Explanation / Answer

Solution :-

a) Given

The laser optical cavity is closed, on one side, by a perfect reflecting mirror and on other side by a partially reflecting mirror.

Also give the later mirro has a smaller transmission coefficient T.

Let = coefficient per unit length.

L = Length of the cavity.

We know that

is directly proportional to T/2L

Hence = c (T/2L) - Equation 1

Also we know that = 2- Equation 2

From Equation 1 and Equation 2, we get

= c ( 2+T/2L)

Hence proved.

b) If   =0

then = c ( 0+T/2L) = = c (T/2L)

It means Iout (T) is directly proprtional to T

Therefore when T--> 0

Then Iout(T) =TI(T)

Hence proved.

This relation is not physiaclly acceptable because in this case it is independent of length of the cavity L.

c) Consider << T/2L.

Hence in this case factor T/2L can be ignored

Hence T^2 = (2L)^2 * R considering Iout(T) maximum value

Therefore

T = (2L)* Sqrt(R)'

Hence proved.

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