Question

Consider a ruby crystal with two energy levels separated by an energy difference corresponding to a free-space wavelength lambda_0 = 694.3nm, with a Lorentzian lineshape of width Delta v = 330GHz. The spontaneous lifetime is t_sp = 3ms and the refractive index of ruby is n=1.76. What value should the population difference N be to achieve a gain coefficient gamma(v_0) = 0.5 cm^-1 ? How long should the crystal be to provide an overall gain of 4 at the central frequency when gamma(v_0) = 0.5 cm^-1 ? The gain coefficient at central wavelength for Lorentzian lineshape is given as: gamma (v_0) = N(lambda_0/2 pi n)^2 middot 1/t_sp Delta v

Answer 1

a. 1.2557 × 10cm⁻³

b. 0.5050 cm

Step-by-step explanation:

`(N_2)/(N_1) = exp(hc/\lambda_0 kT) = 9.5511 * 10^(-31) << 1`

N ≅
`-N_(\alpha)`

`g(v_0) = (4)/(2 \pi \delta v)= 1.93 * 10^(-12)Hz^(-1)`

`\lambda = (\lambda_0)/(n) = 394.48863 nm`

`\alpha(v_0) = -N_(\alpha) * (\lambda^2)/(8\pi t_(sp))g(v)=-2190cm^(-1)`

a.

`y(v) = N\sigma (v) = N(\lambda^2)/(8\pi t_(sp))g(v)`

`N = (0.5 * 10)/((\lambda^2)/(8\pi t_(sp))g(v_0)) = 1.2557 * 10^(19)`

`G(v) = exp(y(v)z) = 4`