Laser Physics I (PHYS/ECE 464), Fall 2020 Homework #8, Due Wednesday Oct. 28
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Laser Physics I (PHYS/ECE 464), Fall 2020
Homework #8, Due Wednesday Oct. 28
(a) For this problem, the key is to recall that when the laser is oscillating, the upper state population
N2 , and thus the gain saturates to loss levels. When the cavity is blocked, N2 increases back to
its unsaturated (small signal) value.
Thus
𝑁2 ≈𝑁20
1 + 𝐼𝜈/𝐼𝑠
where 𝐼𝜈 is the total intracavity intensity.
The side fluorescence (spont. Emission) power (from a gain volume V) is
𝑃 = 𝐴21𝑁2ℎ𝜈𝑉
The ratio of fluorescence power in two cases of lasing on and off is therefore:
𝑃
𝑃0=𝑁2
𝑁20 =
1
1 + 𝐼𝜈/𝐼𝑠
(b) 𝑃
𝑃0=
1
2 leading to
𝐼𝜈
𝐼𝑠= 1
(Also remember in a high-Q cavity, output intensity would have been 𝐼𝑜𝑢𝑡 ≈𝑇2𝐼𝜈
2 , in case one was
interested!).
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