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I am aware of the derivation of the Fresnel reflection and transmission coefficients for a single plane wave incident on a planar interface between two dielectrics. So I wondered what happens if there are two plane waves of the same frequency incident on a planar interface from opposite sides, with different angles of incidence, but in the same plane of incidence. I was not sure whether reflection and transmission of each wave can be treated independently, given by the same Fresnel coefficients as for a single incident wave. If this is indeed the case, can someone give a rigorous explanation/derivation?

Since I was not sure about this, I tried to derive the resulting electric fields from first principles, taking into account the usual interface conditions. Assume an EM wave 1 is incident with frequency $\omega$ in medium with refractive index $n_1$ from the left on the interface and wave 2 with frequency $\omega$ in medium with refractive index $n_2$from the right onto the interface. The angle of incidence for wave 1 is $\theta_{i1}$ and for wave 2 it is $\theta_{i2}$. I thought it would be natural to assume that there will be six waves propagating overall in this situation: The two incident waves, two waves propagating in the -z direction on the left side arising from reflection of wave 1 and transmission of wave 2 and two waves propagating in the +z direction on the right side arising from transmission of wave 1 and reflection of wave 2. The coordinate system is set up such that the +z axis extends from left to right normal to the interface, the +y axis is "up" in the plane of incidence and the +x axis points into the page. For the wavevectors of these waves I get:

$$ \mathbf{k}_{i1} = k_{i1} (\sin\theta_{i1} \,\mathbf{\hat{y}}+ \cos\theta_{i1} \,\mathbf{\hat{z}})\\ \mathbf{k}_{r1} = k_{r1} (\sin\theta_{r1} \,\mathbf{\hat{y}}- \cos\theta_{r1} \,\mathbf{\hat{z}})\\ \mathbf{k}_{t1} = k_{t1} (\sin\theta_{t1} \,\mathbf{\hat{y}}+ \cos\theta_{t1} \,\mathbf{\hat{z}}) $$ and $$ \mathbf{k}_{i2} = k_{i2} (\sin\theta_{i2} \,\mathbf{\hat{y}}- \cos\theta_{i2} \,\mathbf{\hat{z}})\\ \mathbf{k}_{r2} = k_{r2} (\sin\theta_{r2} \,\mathbf{\hat{y}}+ \cos\theta_{r2} \,\mathbf{\hat{z}})\\ \mathbf{k}_{t1} = k_{t2} (\sin\theta_{t2} \,\mathbf{\hat{y}}- \cos\theta_{t2} \,\mathbf{\hat{z}}) $$

If I then resolve the electric fields of the incident waves into a p-component (in the plane of incidence) and s-component (normal to the plane of incidence) and require that the parallel component of the electric field (so x- and y-components in my coordinate system) is continuous across the boundary where z=0, I arrive at the following equation:

$$ [E_{i1}^p \cos\theta_{i1} \,\mathbf{\hat{y}} + E_{i1}^s \,\mathbf{\hat{x}}] \,e^{i(k_{i1}y\sin\theta_{i1} - \omega t)} + [E_{r1}^p \cos\theta_{r1} \,\mathbf{\hat{y}} + E_{r1}^s \,\mathbf{\hat{x}}] \,e^{i(k_{r1}y\sin\theta_{r1} - \omega_{r1} t)} + [E_{t2}^p \cos\theta_{t2} \,\mathbf{\hat{y}} + E_{t2}^s \,\mathbf{\hat{x}}] \,e^{i(k_{t2}y\sin\theta_{t2} - \omega_{t2} t)} = \\ [E_{i2}^p \cos\theta_{i2} \,\mathbf{\hat{y}} + E_{i2}^s \,\mathbf{\hat{x}}] \,e^{i(k_{i2}y\sin\theta_{i2} - \omega t)} + [E_{r2}^p \cos\theta_{r2} \,\mathbf{\hat{y}} + E_{r2}^s \,\mathbf{\hat{x}}] \,e^{i(k_{r2}y\sin\theta_{r2} - \omega_{r2} t)} + [E_{t1}^p \cos\theta_{t1} \,\mathbf{\hat{y}} + E_{t1}^s \,\mathbf{\hat{x}}] \,e^{i(k_{t1}y\sin\theta_{t1} - \omega_{t1} t)} $$

This has to be fulfilled at all times and at all points on the interface, so for all y. I assumed this requires that all the exponentials are equal, in which case it follows that

$$ \omega_{r1} = \omega_{t1} = \omega_{r2} = \omega_{t2} = \omega $$ as one might expect. However, it also leads to

$$ k_{i1}\sin\theta_{i1} = k_{i2}\sin\theta_{i2} $$ which cannot be satisfied in general because I do not require the angles of incidence for the two waves to be the same and the k are fixed since $k_{1, 2} =\frac{n_{1, 2} \omega}{c}$. So the boundary conditions would impose a constraint that cannot be satisfied in general. How can this discrepancy be resolved?

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The problem is linear so it reduces to two independent problems, each which can be solved using Fresnel's equations.

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  • $\begingroup$ Could you elaborate on how mathematically the linearity of Maxwell's equations leads to a separation of the two waves in independent equations? $\endgroup$ – Quantum Jul 14 '18 at 23:10

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