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?


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

  • $\begingroup$ 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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