Intensity conservation during refraction

[lakmus]

Hi,
I try to solve excercises from the book Stellar Atmospheres by Mihallas. I'm stuck on this one:
By use of Snell's law,
$n_1(\nu)\sin{(\theta_1)}=n_2(\nu)\sin{(\theta_2)}$,
in the calculation of the energy passing through an unit area on the interface between two dispersive media with differing indices of refraction, show that
$I_{\nu}n_{\nu}^{-2}$
is a constant.

Well I tried at firts write just conservation of energy for intensity (here is the assumption that all light goes thru the interface). From definition of intensity and rewriting and solid angle by
$\mathrm{d}\omega=\frac{\mathrm{d}S \cos{\theta}}{r^2}$
I got:

$I_1 \cos^2{(\theta_1)} = I_2 \cos^2{(\theta_2)}$,
where index 1 is for light in the medium with refraction index $n_1$, index 2 analogically.
Using Snell's law does not helped to get right result.

So I tried calculate some reflection, but I could not get the right result like Mihallas - $I_{\nu}n_{\nu}^{-2}$ .

It's the firt exercise in the whole book, so I supose it should be easy, bud I can't find out, where I'm wrong.
Thank's for all advices!

 

[lippyka]

The correct approach is to use the conservation of energy. Since the light is passing through an interface, the total energy must be preserved. From the definition of intensity, we can calculate the total energy of light in medium 1 and medium 2: I_1 n_1^2 = I_2 n_2^2 where n_1 and n_2 are the indices of refraction for the two media. Using Snell's law, we can rewrite this equation as: I_1 n_1^2 (\sin{\theta_1})^2 = I_2 n_2^2 (\sin{\theta_2})^2 Since sin{\theta_1} = sin{\theta_2}, we can thus rewrite the equation as: I_1 n_1^2 = I_2 n_2^2 which is the same as our original equation. Therefore, we can conclude that I_{\nu}n_{\nu}^{-2} is a constant.