Question

Optoelectronics and Photonics: Principles and Practices 2nd edition Chapter 1.19

1.19 Dielectrie mirror and AlAs with nr = 2.912, both around i 500 nm. The GaAs-AlAs dielectric mirror is inside a vertical cavity surface emitting laser diode operating at 1.5 μ. The substrate is GaAs with n.-n, n"-3382. The light is incident on the mirror from another semiconductor that is GaAlAs with an index nn 3.40. Calculate how many pairs of layers N would be needed to get a reflectance above 95%, what would be the bandwidth? Consider a dielectric mirror that is made up of quarter wave layers of GaAs with Ay 3.382

Answer 1

R_N = [n^2N_1 - (n_0/n_3) n^2N_2/n^2N_1 + (n_0/n_3) n^2N_2] = [(n_1/n_2)^2N - (n_0/n_3)/(n_1/n-2)^2N + (n_0/n_3)]^2 Therefore (n_1/n_2)^2N = (n_0/n_3) 1 + squareroot R_N/1 - R squareroot N Insert R_N = 0.95 and solve, N = 1/2 ln[n_0/n_3 1 + squareroot R_N/1 - squareroot R_N]/ln(n_1/n_2) = 14.58 i.e 15 pairs are needed Delta lambda.lambda_0 asymp 4/pi arcsin (n-1 - n_2/n_1 + n_2) = 0.095167 But lambda_0 = 1500 nm Therefore Delta lambda = 142 nm