6 4 2 0 0 4 5 Photon energy ev Figure 32 Real and negative i

6 4 2 0 0 4 5 Photon energy, ev Figure 3.2. Real and (negative) imaginary parts of the re- fractive index for silicon. [After H. R. Phillip and E. A. Taft, Physical Review 120 (1960), 37-38.1

Solution

Answer:

(a) For wavelength = 1000 nm

Energy = 1242.3/1000 = 1.24 eV, n = 4.5, k ~ 4.5

Reflectance R = (n-1)² + k²/ (n+1)² + k² = (3.5 -1)² + 0/ (3.5 +1)² + 0 = (2.5 x 2.5 )/ (4.5 x 4.5) = 0.30

(b) For wavelength = 300 nm

Energy = 1242.3/300 = 4.14 eV, n = 4.5, k ~ 4.5

Reflectance R = (n-1)² + k²/ (n+1)² + k² = (4.5 -1)² + 4.5² / (4.5 +1)² + 4.5² = 32.5/50.5 = 0.64

(c) For wavelength = 400 nm

Energy = 1242.3/400 = 3.10 eV, n = 5, k ~ 0.5

Reflectance R = (n-1)² + k²/ (n+1)² + k² = (5 -1)² + 0.5² / (5 +1)² + 0.5² = 16.25/36.25 =0.44