Reflection and Transmission of Electromagnetic Waves: Part 2
Looking Back
Given below is a link to the first article in this series where I discussed how to derive the amplitudes of reflected and transmitted waves at a dielectric interface. The first part talked about the normal incidence condition, and I urge you to check it out first since the introduction will come in useful.
If you recall, the case of normal incidence was quite easy to solve. However, when the waves impinge upon the interface at an angle, the mathematics becomes a little bit lengthy. But understanding it is fairly simple. Let us start with the diagram as before.
There are two polarization states possible in this scenario. One where the electric wave vector is polarized perpendicular to the plane of incidence, and one where the magnetic field vector is polarized perpendicular to the plane of incidence. It might be prudent to mention here what I mean by the term “plane of incidence”. As we will soon derive, the incident, reflected, and transmitted waves lie in the same plane, and I define it as follows:
The plane which contains the surface normal and the propagation vector of the incident radiation is known as the plane of incidence. Note that this is not the plane containing the interface.
For the case we are discussing, the plane of incidence is the X-Z plane, while the boundary is the X-Y plane. In the image attached below, the semi-transparent red plane is the incidence plane, while the boundary plane is marked in green.
Looking at the Polarization States
Now, we first turn our attention to the polarization state where the magnetic waves are polarized perpendicular to the incidence plane. Here are how the propagation directions will look like.
The magnetic wave polarization is perpendicular to the incident plane and thus, will always be along the y-axis, either positive or negative. Considering that the diagram I attached in the beginning took the direction of propagation to be along the z-axis, and the dielectric interface to be the X-Y plane, the y-axis points into or out of the screen from your perspective. For the incident and transmitted waves, B will point out of the screen. For the reflected wave, it points into the screen. We now turn our attention to the mathematics of this topic.
The Laws of Reflection and Refraction
As before, I start by expressing all the waves in complex notation. This time, in place of z, we will include the more generic r vector since the waves travel in three dimensions now. Once again, the first medium has permeability μ₁, dielectric constant ε₁ and the second medium is characterized by μ₂ and ε₂. The velocities of the wave are similarly labelled v₁ and v₂. The equations become:
Do you remember what happens when a wave traverses across the boundary of two media? Its speed changes, which is balanced by a change in the wavelength. The frequency, however, remains constant. Thus, we can relate the magnitude of the wavevectors of each of the three waves. Indeed,
kᵢv₁ = kᵣv₁ = kₜv₂ = ω. Thus, kᵢ = kᵣ = (v₂/v₁) kₜ = (n₁/n₂) kₜ. — (7)
There is nothing interesting about that result. It is merely the relation between the incident and transmitted wavelengths. Naturally, the wavelength remains invariant during reflection. Next, we look at the boundary conditions.
Now, regardless of what component of the fields we look at, the generic structure is the same. In each term, we have an amplitude part, which might include a cosine or a sine part if we resolve the wave vectors along or perpendicular to the boundary, and along with this amplitude part, we have an exponential factor. Thus, any boundary condition will have the same structure as given below:
The exponential parts of this equation must be equal simultaneously. This is evident from the fact that there is no spatial dependence in the brackets. The amplitudes do not contain x, y, or z. Thus, without the exponents being equal, there is no way for the above equality to be satisfied. Since the frequency is equal, the temporal parts are already equal, and we are left with the spatial part. That is, kᵢ . r = kᵣ . r = kₜ . r. Given that z = 0 at the interface, this equation is satisfied independently for the x and y parts. Let us represent r and k in terms of their components. We will then have
Using the trigonometric resolution of each of these wave vectors, we can derive a relation between the angles of incidence, reflection, and transmission. Remember that we have already established the equality of kᵢ and kᵣ. Since kᵢ . r = kᵣ . r = kₜ . r, we have:
Note that the wavevectors do not have a component along the y direction in this case. But even if they did, the above equalities would still be satisfied separately for x, y, and z. Use equation 7 to connect each component from the above relations. You will arrive at the following remarkable and famous results:
- Firstly, if the x, y, or z component of kᵢ is zero, then so are the corresponding components of kᵣ and kₜ. Thus, if the incident wave lies in the X-Z plane, then so do the other two waves.
In other words, the incident, reflected, and transmitted waves all lie in one single plane. This is the first law of reflection. - Secondly, sin θᵢ = sin θᵣ and thus, θᵢ = θᵣ. That is, the angle of incidence is the same as the angle of reflection. Note that these angles are measured with respect to the surface normal at the point of incidence. This is known as the second law of reflection.
- And finally, using equation 7 and vᵢ = (c/nᵢ), where n is the refractive index, we have n₁ sin (θᵢ) = n₂ sin (θₜ). This result is known as Snell’s law of refraction.
These three results were a necessary digression from our main discussion. Indeed, you will find that these come in handy as we proceed.
The Boundary Conditions
I’ve reattached the polarization states image and resolved the vectors along and perpendicular to the boundary. Please excuse the mess since three-dimensional art is not one of my fortes. But you can see from the image that the magnetic wave vector has no components perpendicular to the interface. Thus, one boundary condition goes poof! And using (i), (iii), and (iv), we have, respectively.
As before, let β = μ₁v₁/μ₂v₂ = μ₁n₂/μ₂n₁, and α = cos θₜ /cos θᵢ. Then,
These relations between the incident, reflected, and transmitted waves are known as Fresnel’s equations, and play a significant role in optics. The factor α depends on the incident angle and thus, the amplitudes of the reflected and transmitted waves do so as well.
Finishing Up
Once again, we can find the reflection and transmission coefficients, which sum up to 1, proving the conservation of energy. I will leave that as an exercise for you. The coefficients are the ratios of the intensities of reflected or transmitted and incident waves. Note that this time there will be a factor of cos θ in each of the expressions for intensity I, since the power per unit area is given by the dot product S.z. Here, S is the well-known Poynting vector, something that I have already discussed in my article about Conservation Laws in Electrodynamics. You can see it here:
Further, one can derive the angle at which the reflected wave is completely extinguished, known as the Brewster’s angle, and approximated by tan (θᵦ) ≈ n₂/n₁. In the next and final part of this article, I will discuss the scenario where the incident wave vector is polarized perpendicular to the plane of incidence. Until then, keep learning and keep enjoying Physics.
