 Solution

ID: past-exam-of-the-atomic-and-laser-physics-course-of-the-university-of-oxford/2022/4/c/solution

Let

x = I_{0} / I_{s}

and

G = α_{0} L

. Applying the integrated amplifier law to each section gives

ln 70 + 69 x = G, ln \frac{150}{70} + 80 x = G .

Eliminating

G

gives

\frac{I _{0}}{I _{s}} = \frac{1}{11} ln \frac{7 0 ^{2}}{150} ≃ 0.31694.

G ≃ 26.1175

. The second section receives a larger intensity and adds

80 I_{0}

, compared with

69 I_{0}

in the first, while its multiplicative optical gain is much smaller. This is the expected approach to constant extracted power in saturation.

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