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## Homework Statement

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(a) Why does input intensity affect gain?

(b) Derive the expression

(c) Find saturation intensity in terms of input intensity

## Homework Equations

## The Attempt at a Solution

__Part(a)__Consider a narrow band radiation as an input, where its bandwidth is much smaller than the spectral with of transitions, so in general

[tex]\frac{dN_2}{dt} = S_2 - (N_2B_{21}-N_1B_{12}) \int g_H(\omega - \omega_0) \rho(\omega) d\omega + \cdots[/tex]

[tex] = S_2 - N^{*} \int B_{21} g_H(\omega - \omega_0)\rho(\omega) d\omega + \cdots[/tex]

[tex] = S_2 - N^{*}\sigma_{21}(\omega_L - \omega_0) \frac{I}{\hbar \omega_L}[/tex]

Thus, we see that rates depend on input intensity ##I_T##, which influences the gain on a laser amplifier when we solve for the steady state solutions.

__Part(b)__Bookwork. Managed to derive it.

[tex]\alpha(\omega) = \frac{\alpha_0(\omega)}{1 + \frac{I}{I_{sat}}} [/tex]

where saturation intensity is ## I_{sat} = \frac{\hbar \omega_L}{\sigma_{21} \tau_R}## and relaxation time is ##\tau_R = \tau_2 + \frac{g_2}{g_1} \tau_1 (1- \tau_2A_{21})##.

__Part(c)__The equation for intensity is given by

[tex]\frac{dI}{dz} = \alpha I = \frac{\alpha_0}{1 + \frac{I}{I_{sat}}}I[/tex]

which may be integrated to give

[tex] ln \left( \frac{I(z)}{I_{(0)}} \right) + \frac{I_{(z)} - I_{(0)}}{I_{sat}} = \alpha_0 z [/tex]

At low intensity where ##I_{(z)} << I_{sat}##, the equation becomes ##I_{(z)} = I_{(0)} = e^{\alpha_0 z}##.

At high intensity where ##I_{(z)} \approx I_{(0)}##, the equation becomes ##I_{(z)} = I_{(0)} + \alpha_0 I_{sat} z##.

Thus initially the beam intensity is weak, then it becomes strong, so

[tex]e^{\alpha_0 z} = 100[/tex]

[tex] 200I_0 = \alpha_0 I_{sat}z[/tex]