$ \lambda = 633nm\hspace{0.5cm}n_2=1.52 $

a)

$ r=\frac{n_1-n_2}{n_1+n_2} = \frac{1-1.52}{1+1.52} = \frac{-0.52}{2.52} = \frac{-52}{252} $

$ r = \frac{-13}{63} $

$ P = \frac{1}{2n_0}|E|^2 $

$ \to R = r^2 = \frac{169}{3969}\approx \frac{17}{400} \approx \frac{1}{24} $


b) $ n_1 = \sqrt{1.52} > 1 \to $ realistic refractive index

c) $ d = \frac{\lambda}{4} = \frac{633\cdot 10^{-9}m}{4} = 158.25\cdot10^{-9}m $

d) $ \Gamma = \frac{n_2-n_1^2}{n_2+n_1^2} = \frac{1.52-1.38^2}{1.52 +1.38^2}\approx = \frac{0.44}{3.48} = \frac{11}{87} \approx\frac{1}{8}\to\frac{1}{64} $

$ r^2-\Gamma^2 = \frac{1}{24}-\frac{1}{64} = \frac{64-24}{1536} = \frac{40}{1536}\approx \frac{1}{40} \text{ less power reflected} $

Alumni Liaison

BSEE 2004, current Ph.D. student researching signal and image processing.

Landis Huffman