Put
$$
k=\frac{\omega}{c_0},\qquad
p=k\sin\theta_i,\qquad
q_0=k\cos\theta_i,
$$
and let $q_1=nk\cos\theta_T$ be the downward <vertical wavenumber> in the lower half-space. The given form of <Snell's law> says
$$
q_1=q_0\sqrt{1+\alpha}.
$$
With the $e^{-i\omega t}$ convention, write the incident, reflected, and transmitted <plane waves> as
$$
\psi_i=e^{i(px-q_0z)},\qquad
\psi_r=R e^{i(px+q_0z)},\qquad
\psi_t=T e^{i(px-q_1z)}.
$$
The two <interface condition>[interface conditions] at $z=0$ give
$$
1+R=T,
\qquad
q_0(1-R)=q_1T.
$$
Solving this <linear system> gives the <reflection and transmission coefficients at a scalar-wave interface>
$$
\boxed{R=\frac{q_0-q_1}{q_0+q_1}
=\frac{1-\sqrt{1+\alpha}}{1+\sqrt{1+\alpha}}},
\qquad
\boxed{T=\frac{2q_0}{q_0+q_1}
=\frac{2}{1+\sqrt{1+\alpha}}}.
$$
Thus the exact fields are
$$
\boxed{\psi_0=e^{i(px-q_0z)}+R e^{i(px+q_0z)}}
\quad(z>0),
$$
$$
\boxed{\psi_1=T e^{i(px-q_1z)}}
\quad(z<0).
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 335 / 1 / i / Solution

{title2=$q$}

Relative to a distinguished propagation coordinate $z$, the vertical wavenumber is the corresponding wavevector component $q$. For a plane wave of wavenumber $k$ making angle $\theta$ with the vertical, $q=k\cos\theta$.


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