 Quote by DrDu
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.
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Yes, I'm pretty sure this is the [itex]θ'[/itex] I'm talking about and I think we're on the same wavelength(/vector... ohh, puns).
But when you say:
Quote by DrDu
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.
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Please tell me if I'm interpreting this incorrectly, but is this a more physically rigorous way of implying that the real component of the angle [itex]θ'[/itex] must be zero? i.e. [itex]θ'[/itex] is pure-imaginary? This is the result when I take Pi/2 - [itex]θ_{t}[/itex], and it does seem to make some intuitive sense.