Refractive Index: Homework Answers

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SUMMARY

The discussion centers on solving a homework problem related to the refractive index using the equation ## \frac { \sin \theta } {\sin \theta ' } = \frac { n_2} { n_1} ##. The user establishes that the dielectric constant for air is 1, resulting in ## n_1 = 1 ##. For the second medium, they derive the equation for the refractive index as ## n_2 = \sqrt{ \frac { E } { E-V_0 } } = 1+ \frac { V_0 } {2E } ##. The user suspects that the Schrödinger equation is relevant to the problem, particularly in the context of an incident plane wave and potential reflections at the boundary.

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  • Explore the Schrödinger equation in two dimensions
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Pushoam
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Homework Statement


upload_2017-12-12_14-9-3.png


Homework Equations

The Attempt at a Solution


I think that ## \frac { \sin \theta } {\sin \theta ' } = \frac { n_2} { n_1} ##

I am taking the dielectric constant as 1 as the potential is 0 i.e. the medium is air..

So, ##n_1 = 1##

For the 2nd medium,

The dielectric constant ## \epsilon_r = \frac { E } { E- V_0} ##

So, ## n_2 = \sqrt{ \frac { E } { E-V_0 } } = 1+ \frac { V_0 } {2E } ##

What is wrong here?
 

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I haven't solved it in its entirety yet, but I think the problem makes use of the Schrödinger equation. Try writing out the Schrödinger equation in two dimensions with an incident plane wave particle with ## \psi (\vec{r})=e^{i \vec{k}_o \cdot \vec{r} } ## and ## \vec{k}_o=k_{xo}\hat{i}+k_{yo} \hat{j} ##. ## \\ ## Editing: To complicate the problem, you might also get a partial reflection at the boundary, with angle of incidence =angle of reflection. (I'm not sure because I didn't solve it completely yet).
 
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