| Line 2: | Line 2: | ||
a) | a) | ||
| + | |||
<math>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}</math> | <math>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}</math> | ||
Revision as of 12:32, 4 June 2017
$ \lambda = 633nm\hspace{0.5cm}n_2=1.52<math> a) <math>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} $
