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

Table of contents
- Solution i

Solution

Put

k = \frac{ω}{c _{0}}, p = k sin θ_{i}, q_{0} = k cos θ_{i},

and let

q_{1} = nk cos θ_{T}

be the downward vertical wavenumber in the lower half-space. The given form of Snell's law says

q_{1} = q_{0} 1 + α .

With the

e^{- iω t}

convention, write the incident, reflected, and transmitted plane waves as

ψ_{i} = e^{i (p x - q_{0} z)}, ψ_{r} = R e^{i (p x + q_{0} z)}, ψ_{t} = T e^{i (p x - q_{1} z)} .

The two interface conditions at

z = 0

give

1 + R = T, q_{0} (1 - R) = q_{1} T .

Solving this linear system gives the reflection and transmission coefficients at a scalar-wave interface

R = \frac{q _{0} - q _{1}}{q _{0} + q _{1}} = \frac{1 - 1 + α}{1 + 1 + α}, T = \frac{2 q _{0}}{q _{0} + q _{1}} = \frac{2}{1 + 1 + α} .

Thus the exact fields are

ψ_{0} = e^{i (p x - q_{0} z)} + R e^{i (p x + q_{0} z)} (z > 0),

ψ_{1} = T e^{i (p x - q_{1} z)} (z < 0) .

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