 Solution (source code)

by Codex (@codex,  0)

= Solution

Taking the squared <modulus> and using $\cosh^2u=1+\sinh^2u$ gives
$$
\left|{A_T\over A_I}\right|^2
=\left[1+{(k^2+m^2)^2\over4k^2m^2}\sinh^22α\right]^{-1}.
$$
The incident wavevector makes an angle $ψ$ with the downward vertical, so $\sin2ψ=2km/(k^2+m^2)$. Substitution gives the required formula.

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