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

Answer:

a. 1.2557 × 10cm⁻³

b. 0.5050 cm

Explanation:

$\frac{N_2}{N_1} = exp(hc/\lambda_0 kT) = 9.5511 \times 10^{-31} << 1$

N ≅ $-N_{\alpha}$

$g(v_0) = \frac{4}{2 \pi \delta v}= 1.93 \times 10^{-12}Hz^{-1}$

$\lambda = \frac{\lambda_0}{n} = 394.48863 nm$

$\alpha(v_0) = -N_{\alpha} \times \frac{\lambda^2}{8\pi t_{sp}}g(v)=-2190cm^{-1}$

a.

$y(v) = N\sigma (v) = N\frac{\lambda^2}{8\pi t_{sp}}g(v)$

$N = \frac{0.5 \times 10}{\frac{\lambda^2}{8\pi t_{sp}}g(v_0)} = 1.2557 \times 10^{19}$

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