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.

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.

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 pureimaginary? This is the result when I take Pi/2  [itex]θ_{t}[/itex], and it does seem to make some intuitive sense.