Normal Incidence: Reflection & Transmission at Dielectric Interfaces

Hello! Welcome to the fourth lesson in our module on Electromagnetic Waves and Optics.

In our last lesson, we established how to quantify the energy flow of an electromagnetic wave using the Poynting vector ( $S$ ) and its time-averaged magnitude, the intensity ( $I$ ). We saw that $I = c ⟨ u ⟩$ , meaning the intensity is the average energy density moving at the speed of light.

Today, we address the fundamental question of what happens when this energy encounters a material boundary. Our learning outcome is to analyze the reflection and transmission of electromagnetic waves at normal incidence on a dielectric interface. This is the simplest, yet foundational, case for understanding how light interacts with materials like glass, water, or optical fibers. We will derive the coefficients that tell us exactly what fraction of the wave is reflected and what fraction passes through.

1. The Scenario: Normal Incidence

Let's set up the problem. Imagine a plane electromagnetic wave, traveling in the +z direction through a medium 1 (e.g., air), which has permittivity $ϵ_{1}$ , permeability $μ_{1}$ , and refractive index $n_{1}$ . At the plane $z = 0$ , it encounters a second medium (e.g., glass) with properties $ϵ_{2}, μ_{2}, n_{2}$ . The wave strikes the interface at a 90° angle, which we call normal incidence.

Upon hitting the boundary, part of the wave reflects back into medium 1 (traveling in the -z direction), and part is transmitted into medium 2 (continuing in the +z direction).

Reflection and Transmission of a Plane EM Wave at Normal Incidence — Caption: A plane wave, with fields $\mathbf{E}_i, \mathbf{H}_i$, is normally incident on a dielectric interface at z=0. This results in a reflected wave ($\mathbf{E}_r, \mathbf{H}_r$) and a transmitted wave ($\mathbf{E}_t, \mathbf{H}_t$). The fields are shown for a wave polarized in the x-direction.

We can describe these three waves using the complex exponential form we've studied previously. Let's assume the incident electric field is polarized along the x-axis:

Incident Wave ( $z < 0$ ):
$E_{i} (z, t) = x^E_{0 i} e^{i (k_{1} z - ω t)}$
$B_{i} (z, t) = y^\frac{E _{0 i}}{v _{1}} e^{i (k_{1} z - ω t)}$
Reflected Wave ( $z < 0$ ):
$E_{r} (z, t) = x^E_{0 r} e^{i (- k_{1} z - ω t)}$
$B_{r} (z, t) = - y^\frac{E _{0 r}}{v _{1}} e^{i (- k_{1} z - ω t)}$
Transmitted Wave ( $z > 0$ ):
$E_{t} (z, t) = x^E_{0 t} e^{i (k_{2} z - ω t)}$
$B_{t} (z, t) = y^\frac{E _{0 t}}{v _{2}} e^{i (k_{2} z - ω t)}$

Notice the sign change in the reflected magnetic field. This is required for the reflected wave to remain a transverse wave where $E$ , $B$ , and the propagation vector $k$ form a right-handed system ( $S \propto E \times B$ ). Since $k_{r}$ points in the -z direction, if $E_{r}$ is in the +x direction, $B_{r}$ must be in the -y direction.

Our goal is to find the reflection coefficient ( $r$ ) and transmission coefficient ( $t$ ), which relate the amplitudes of the reflected and transmitted waves to the incident wave:
$r = \frac{E _{0 r}}{E _{0 i}} and t = \frac{E _{0 t}}{E _{0 i}}$

2. Applying Electromagnetic Boundary Conditions

To solve for $r$ and $t$ , we use the fundamental boundary conditions that govern electromagnetic fields at an interface between two dielectric media (assuming no free charges or surface currents). These conditions state that the components of the $E$ and $H$ fields parallel (tangential) to the boundary must be continuous.

Since you appreciate mathematical rigor, the following reading from LibreTexts shows how these boundary conditions are formally derived from the integral form of Maxwell's equations.

Reflection and Transmission at an Interface

This resource provides a rigorous derivation of the boundary conditions for electromagnetic fields at an interface using Stokes' and Gauss's theorems. Understanding this foundation is key to confidently applying these rules.

Please read the subsection '1.9.1 Boundary Conditions at an Interface'. Focus on how integrating Maxwell's equations over a small loop or pillbox straddling the boundary leads to the continuity relations for the tangential and normal field components. You don't need to read the whole resource, just this specific subsection.

For our normal incidence case, the $E$ and $H$ fields are entirely tangential to the boundary plane ( $z = 0$ ). The total field in region 1 is the sum of the incident and reflected fields.

Continuity of tangential $E$ :
$E_{i} (0, t) + E_{r} (0, t) = E_{t} (0, t)$
At $z = 0$ , the exponential terms $e^{- iω t}$ are common to all fields and cancel out, leaving a relationship between the amplitudes:
$E_{0 i} + E_{0 r} = E_{0 t} (Eq. 1)$
Continuity of tangential $H = B / μ$ :
$\frac{1}{μ _{1}} B_{i} (0, t) + \frac{1}{μ _{1}} B_{r} (0, t) = \frac{1}{μ _{2}} B_{t} (0, t)$
Substituting the field amplitudes and canceling the time dependence gives:
$\frac{1}{μ _{1}} (\frac{E _{0 i}}{v _{1}} - \frac{E _{0 r}}{v _{1}}) = \frac{1}{μ _{2}} \frac{E _{0 t}}{v _{2}} (Eq. 2)$

3. Deriving the Amplitude Coefficients

We now have a system of two equations with two unknowns ( $E_{0 r}$ and $E_{0 t}$ ). Let's solve them. For most optical materials, the permeability is very close to that of free space ( $μ_{1} \approx μ_{2} \approx μ_{0}$ ). Under this common approximation, Eq. 2 simplifies. Recalling that the wave speed in a medium is $v = c / n$ , we get:
$n_{1} (E_{0 i} - E_{0 r}) = n_{2} E_{0 t} (Eq. 2a)$

The following video provides an excellent walkthrough of the setup and the algebraic solution of these two equations.

Fresnel Equations at Normal Incidence

This video, by Jordan Edmunds, clearly explains how to set up the fields and apply the boundary conditions to derive the Fresnel equations for normal incidence. It's a very practical and clear demonstration of the physics we've just discussed.

Watch the video from the beginning up to the 11:08 mark. Pay attention to how the field directions are established, how the boundary conditions are written down, and how the resulting simultaneous equations are solved to obtain the reflection and transmission coefficients.

Let's summarize the derivation you saw in the video. We substitute Eq. 1 into Eq. 2a:
$n_{1} (E_{0 i} - E_{0 r}) = n_{2} (E_{0 i} + E_{0 r})$
Now, we solve for the ratio $E_{0 r} / E_{0 i}$ :
$n_{1} E_{0 i} - n_{1} E_{0 r} = n_{2} E_{0 i} + n_{2} E_{0 r}$
$(n_{1} - n_{2}) E_{0 i} = (n_{1} + n_{2}) E_{0 r}$
This gives the amplitude reflection coefficient $r$ :
$r = \frac{E _{0 r}}{E _{0 i}} = \frac{n _{1} - n _{2}}{n _{1} + n _{2}}$
To find the amplitude transmission coefficient $t$ , we can substitute $E_{0 r} = r E_{0 i}$ back into Eq. 1:
$E_{0 i} + r E_{0 i} = E_{0 t} ⟹ t = \frac{E _{0 t}}{E _{0 i}} = 1 + r$
$t = 1 + \frac{n _{1} - n _{2}}{n _{1} + n _{2}} = \frac{( n _{1} + n _{2} ) + ( n _{1} - n _{2} )}{n _{1} + n _{2}} = \frac{2 n _{1}}{n _{1} + n _{2}}$

A crucial point: A negative value for $r$ (which happens when $n_{2} > n_{1}$ ) signifies a 180° phase shift of the electric field upon reflection. The reflected E-field points in the opposite direction to what we initially drew. This is a famous phenomenon in optics.

4. Reflectance and Transmittance: The Flow of Energy

The amplitude coefficients $r$ and $t$ tell us about the fields, but what about the energy? We are typically interested in the fraction of incident power or intensity that is reflected and transmitted. These are given by the Reflectance ( $R$ ) and Transmittance ( $T$ ).

From our last lesson, we know intensity is proportional to the square of the electric field amplitude: $I = \frac{1}{2} v ϵ E_{0}^{2} = \frac{1}{2} \frac{n}{c μ _{0}} E_{0}^{2}$ .

The Reflectance $R$ is the ratio of reflected intensity to incident intensity:
$R = \frac{I _{r}}{I _{i}} = \frac{\frac{1}{2} v _{1} ϵ _{1} E _{0 r}^{2}}{\frac{1}{2} v _{1} ϵ _{1} E _{0 i}^{2}} = (\frac{E _{0 r}}{E _{0 i}})^{2} = r^{2}$
$R = (\frac{n _{1} - n _{2}}{n _{1} + n _{2}})^{2}$

The Transmittance $T$ is the ratio of transmitted intensity to incident intensity. Be careful here, as the medium changes:
$T = \frac{I _{t}}{I _{i}} = \frac{\frac{1}{2} v _{2} ϵ _{2} E _{0 t}^{2}}{\frac{1}{2} v _{1} ϵ _{1} E _{0 i}^{2}} = \frac{v _{2} ϵ _{2}}{v _{1} ϵ _{1}} (\frac{E _{0 t}}{E _{0 i}})^{2} = \frac{n _{2}}{n _{1}} t^{2}$
$T = \frac{n _{2}}{n _{1}} (\frac{2 n _{1}}{n _{1} + n _{2}})^{2} = \frac{4 n _{1} n _{2}}{( n _{1} + n _{2} ) ^{2}}$

For these lossless dielectrics, energy must be conserved. Let's check if the incident power equals the sum of the reflected and transmitted power:
$R + T = (\frac{n _{1} - n _{2}}{n _{1} + n _{2}})^{2} + \frac{4 n _{1} n _{2}}{( n _{1} + n _{2} ) ^{2}} = \frac{n _{1}^{2} - 2 n _{1} n _{2} + n _{2}^{2} + 4 n _{1} n _{2}}{( n _{1} + n _{2} ) ^{2}} = \frac{n _{1}^{2} + 2 n _{1} n _{2} + n _{2}^{2}}{( n _{1} + n _{2} ) ^{2}} = \frac{( n _{1} + n _{2} ) ^{2}}{( n _{1} + n _{2} ) ^{2}} = 1$
Indeed, $R + T = 1$ , confirming conservation of energy.

The following resource provides a concise summary of this power-flow analysis.

Normal Incidence onto a Dielectric

This article from LibreTexts focuses directly on the power flow analysis for normal incidence, confirming the energy conservation relation R+T=1.

Please read the subsection 'Time-Average Power Flow'. It explicitly shows that the net power flowing into the second medium is the incident power minus the reflected power, and verifies that this equals the transmitted power.

Test your understanding!

A beam of light in air ( $n_{1} \approx 1.00$ ) is normally incident on the surface of a pond of water ( $n_{2} \approx 1.33$ ).

What percentage of the light intensity is reflected?
Does the reflected electric field undergo a phase shift?
What is the ratio of the magnetic field amplitude of the transmitted wave to the incident wave ( $B_{0 t} / B_{0 i}$ )?

Show answer

Reflected Intensity: The percentage of reflected intensity is given by the Reflectance, $R$ .
$R = (\frac{n _{1} - n _{2}}{n _{1} + n _{2}})^{2} = (\frac{1.00 - 1.33}{1.00 + 1.33})^{2} = (\frac{- 0.33}{2.33})^{2} \approx (- 0.1416)^{2} \approx 0.020$
So, about 2.0% of the light intensity is reflected.
Phase Shift: The amplitude reflection coefficient is $r = \frac{1.00 - 1.33}{1.00 + 1.33} \approx - 0.1416$ . Since $r$ is negative, the reflected electric field is inverted, meaning it undergoes a 180° phase shift.
Magnetic Field Amplitude Ratio: We know $B_{0} = E_{0} / v = n E_{0} / c$ .
$\frac{B _{0 t}}{B _{0 i}} = \frac{n _{2} E _{0 t} / c}{n _{1} E _{0 i} / c} = \frac{n _{2}}{n _{1}} \frac{E _{0 t}}{E _{0 i}} = \frac{n _{2}}{n _{1}} t$
Using the formula for $t$ :
$\frac{B _{0 t}}{B _{0 i}} = \frac{n _{2}}{n _{1}} (\frac{2 n _{1}}{n _{1} + n _{2}}) = \frac{2 n _{2}}{n _{1} + n _{2}}$
Plugging in the numbers:
$\frac{B _{0 t}}{B _{0 i}} = \frac{2 ( 1.33 )}{1.00 + 1.33} = \frac{2.66}{2.33} \approx 1.14$
The magnetic field amplitude is actually larger in the water than in the air for the same incident electric field.

Conclusion

Today we have built a crucial bridge between the theory of electromagnetic waves and the practical phenomena of reflection and transmission. By applying the fundamental boundary conditions of electromagnetism, we derived quantitative expressions for how much of a wave's amplitude and intensity are reflected and transmitted at a normal-incidence boundary.

Key Takeaways:

The behavior of an EM wave at an interface is governed by the continuity of tangential $E$ and $H$ fields.
For normal incidence on a dielectric, the amplitude reflection ( $r$ ) and transmission ( $t$ ) coefficients depend on the refractive indices of the two media:
$r = \frac{n _{1} - n _{2}}{n _{1} + n _{2}} and t = \frac{2 n _{1}}{n _{1} + n _{2}}$
A negative reflection coefficient ( $r < 0$ ), which occurs when reflecting off a medium with a higher refractive index ( $n_{2} > n_{1}$ ), corresponds to a 180° phase shift in the electric field.
The Reflectance ( $R$ ) and Transmittance ( $T$ ), which represent the fractions of reflected and transmitted intensity, are given by:
$R = r^{2} and T = \frac{n _{2}}{n _{1}} t^{2}$
Energy is conserved in this interaction, as demonstrated by $R + T = 1$ .

Preview of the Next Lesson:
Normal incidence is a special, highly symmetric case. What happens if the wave strikes the interface at an angle? The physics becomes richer, involving polarization and phenomena like the Brewster angle. In our next lesson, we will generalize our analysis to derive the Fresnel equations for reflection and transmission at oblique incidence, which govern the behavior of light in almost all real-world optical scenarios.

Recommended resources

1. The Scenario: Normal Incidence

2. Applying Electromagnetic Boundary Conditions

3. Deriving the Amplitude Coefficients

4. Reflectance and Transmittance: The Flow of Energy

Conclusion