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 Author Topic: Intensity of reflected s-wave/p-wave  (Read 39348 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
Fu-Kwun Hwang
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 « Embed this message on: March 17, 2006, 09:22:18 am » posted from:Taipei,T'ai-pei,Taiwan

This applet will calculate reflected and refracted light intensity according to the incident angle for p-wave and s-wave.
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Refraction satisfy Snell's law : $n_1 \sin\theta_1=n_2\sin\theta_2$

There are always reflected wave when light touch the boundary between two media.
The intensity for reflected wave $\vec{E}'= r_s \vec{E}_s + r_p \vec{E}_p$ where $\vec{E}_s, \vec{E}_p$ are electric field for s-wave and p-wave.
$r_s= \frac{n_1\cos\theta_1 -n_2\cos\theta_2}{n_1\cos\theta_1 +n_2\cos\theta_2} =\frac{\sin(\theta_1-\theta_2)}{\sin(\theta_1+\theta_2)}$
and
$r_p=\frac{n_2\cos\theta_1-n_1\cos\theta_2}{n_2\cos\theta_1+n_1\cos\theta_2}=\frac{\tan(\theta_1-\theta_2)}{\tan(\theta_1+\theta_2)}$
The intensity of refraction wave is the intensity of incident wave minus intensity of reflected wave.
There is no refracted wave when internal reflection occurs.
When $\theta=\theta=0$, the fraction of reflected wave is $\frac{(n_1-n_2)^2}{(n_1+n_2)^2}$.
for light from air(n=1) to glass (n=1.5), there are only $\frac{(1-1.5)^2}{(1+1.5)^2}=\frac{0.5^2}{2.5^2}=\frac{1}{25}=4%$ light being reflected.
 emwaveboundary_2_20090116183127.gif (10.54 KB, 629x346 - viewed 721 times.) Logged
minusquare
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 « Embed this message Reply #1 on: October 16, 2007, 12:33:43 pm »

Thanks.
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Fu-Kwun Hwang
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 « Embed this message Reply #2 on: October 16, 2007, 04:37:51 pm »

I just use "send to my email account" method, and open the file from my email message without any problem.
May be there is something wrong with your computer? Try to use another computer to check it again.
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The most important medicine is tender love and care. ..."Mother Teresa(1910-1997, Roman Catholic Missionary, 1979 Nobel Peace Prize)"