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Nov16-12, 09:38 AM
Sci Advisor
P: 3,630
It is still not clear to me what you are after, but the Snell angle is the angle between the wavevector k and the normal vector n, i.e. [itex] \cos(\theta)=\vec{k}/|k|\cdot \hat{n}[/itex].
You are interested in the angle [itex] \cos(\theta')=\vec{k}/|k|\cdot \hat{n'}[/itex]. Where n' is the vector tangential to the surface. But this projection onto the surface has to be equal to the projection of the incident wavevector k_i onto n' as the incident and the transmitted (or evanescent) wave have to be in phase everywhere on the surface.