= Solution
{c}
In a uniform steady amplifier, with no extra propagation losses, the local <optical gain> law gives
$$
\frac{dI}{dz}=\frac{\alpha_0I}{1+I/I_s},\qquad
\left(\frac1I+\frac1{I_s}\right)dI=\alpha_0\,dz.
$$
Integrate between the two faces of length $L$ to obtain
$$
\boxed{\ln\frac{I_{\rm out}}{I_{\rm in}}+\frac{I_{\rm out}-I_{\rm in}}{I_s}=\alpha_0L.}
$$
For $I\ll I_s$, this reduces to exponential amplification. In deep <saturation>, the <intensity> increment approaches $\alpha_0I_sL$: finite pumping limits the extracted power.
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