Reflection & Transmission (Fresnel Equations)

Step 1: Write the incident, reflected, and transmitted waves

Place the interface at $z = 0$. The three waves are:

$$\mathbf{E}_I = E_{0I}\,e^{i(\mathbf{k}_I \cdot \mathbf{r} - \omega_I t)}, \quad \mathbf{E}_R = E_{0R}\,e^{i(\mathbf{k}_R \cdot \mathbf{r} - \omega_R t)}, \quad \mathbf{E}_T = E_{0T}\,e^{i(\mathbf{k}_T \cdot \mathbf{r} - \omega_T t)}$$

Step 2: Apply boundary conditions at the interface

The tangential E must be continuous at $z = 0$ for all $x$, $y$, and $t$:

$$E_{0I}\,e^{i(\mathbf{k}_I \cdot \mathbf{r} - \omega_I t)} + E_{0R}\,e^{i(\mathbf{k}_R \cdot \mathbf{r} - \omega_R t)} = E_{0T}\,e^{i(\mathbf{k}_T \cdot \mathbf{r} - \omega_T t)} \quad \text{at } z=0$$

Step 3: Phase matching in time

For the equation to hold at all times, the time dependence must be the same:

$$\omega_I = \omega_R = \omega_T \equiv \omega$$

The frequency does not change upon reflection or transmission.

Step 4: Phase matching along the interface

At $z = 0$, the spatial phases must match for all $x$ and $y$:

$$k_{Ix}x + k_{Iy}y = k_{Rx}x + k_{Ry}y = k_{Tx}x + k_{Ty}y$$

Step 5: Confine to the plane of incidence

Choose the plane of incidence as the $xz$-plane, so $k_{Iy} = 0$. Then $k_{Ry} = k_{Ty} = 0$ (all three waves lie in the same plane), and:

$$k_I\sin\theta_i = k_R\sin\theta_r = k_T\sin\theta_t$$

Step 6: Derive Snell's law

Since $k_I = k_R = n_1\omega/c$ (same medium) and $k_T = n_2\omega/c$:

$$\frac{n_1\omega}{c}\sin\theta_i = \frac{n_1\omega}{c}\sin\theta_r = \frac{n_2\omega}{c}\sin\theta_t$$

From the first equality: $\theta_i = \theta_r$ (law of reflection).

From the first and third: $n_1\sin\theta_i = n_2\sin\theta_t$ (Snell's law).

Step 1: Write the fields at the interface

Incident: $\mathbf{E}_I = E_I\,\hat{x}$, $\mathbf{B}_I = (E_I/v_1)\,\hat{y}$

Reflected: $\mathbf{E}_R = E_R\,\hat{x}$, $\mathbf{B}_R = -(E_R/v_1)\,\hat{y}$ (propagates in $-\hat{z}$)

Transmitted: $\mathbf{E}_T = E_T\,\hat{x}$, $\mathbf{B}_T = (E_T/v_2)\,\hat{y}$

Step 2: Boundary condition on tangential E

$$E_I + E_R = E_T \quad \text{...(i)}$$

Step 3: Boundary condition on tangential H

For non-magnetic media ($\mu_1 = \mu_2 = \mu_0$), continuity of $H_y = B_y/\mu_0$:

$$\frac{1}{v_1}(E_I - E_R) = \frac{1}{v_2}E_T$$

Using $v = c/n$: $n_1(E_I - E_R) = n_2 E_T$ ...(ii)

Step 4: Solve for the reflection coefficient

From (i): $E_T = E_I + E_R$. Substitute into (ii):

$$n_1(E_I - E_R) = n_2(E_I + E_R)$$

$$n_1 E_I - n_1 E_R = n_2 E_I + n_2 E_R$$

$$E_R(n_1 + n_2) = E_I(n_1 - n_2)$$

$$r \equiv \frac{E_R}{E_I} = \frac{n_1 - n_2}{n_1 + n_2}$$

Step 5: Solve for the transmission coefficient

Substitute back into (i):

$$t \equiv \frac{E_T}{E_I} = 1 + r = 1 + \frac{n_1 - n_2}{n_1 + n_2} = \frac{2n_1}{n_1 + n_2}$$

Step 6: Physical interpretation

If $n_2 > n_1$: $r < 0$ (reflected E flips sign, $180°$ phase shift).

If $n_2 < n_1$: $r > 0$ (no phase shift on reflection).

Note $t > 0$ always (no phase shift on transmission at normal incidence).

Step 1: Set up the geometry

Interface at $z = 0$, plane of incidence is $xz$. The incident wave arrives at angle $\theta_i$. By Snell's law, $n_1\sin\theta_i = n_2\sin\theta_t$.

Step 2: s-polarization (TE) -- E perpendicular to plane of incidence

$\mathbf{E}$ is along $\hat{y}$ for all three waves. Boundary conditions (tangential E and tangential H continuous):

$$E_I + E_R = E_T \quad \text{(tangential E)}$$

$$\frac{n_1}{\mu_1}(E_I - E_R)\cos\theta_i = \frac{n_2}{\mu_2}E_T\cos\theta_t \quad \text{(tangential H)}$$

Step 3: Solve for s-polarization coefficients

For non-magnetic media ($\mu_1 = \mu_2$), substitute $E_T = E_I + E_R$ into the H equation:

$$n_1\cos\theta_i(E_I - E_R) = n_2\cos\theta_t(E_I + E_R)$$

$$r_s = \frac{E_R}{E_I} = \frac{n_1\cos\theta_i - n_2\cos\theta_t}{n_1\cos\theta_i + n_2\cos\theta_t}$$

$$t_s = \frac{E_T}{E_I} = \frac{2n_1\cos\theta_i}{n_1\cos\theta_i + n_2\cos\theta_t}$$

Step 4: p-polarization (TM) -- E parallel to plane of incidence

$\mathbf{E}$ lies in the $xz$-plane. The boundary conditions become:

$$(E_I - E_R)\cos\theta_i = E_T\cos\theta_t \quad \text{(tangential E: x-component)}$$

$$\frac{n_1}{\mu_1}(E_I + E_R) = \frac{n_2}{\mu_2}E_T \quad \text{(tangential H: y-component)}$$

Step 5: Solve for p-polarization coefficients

For non-magnetic media, from the H condition: $E_T = n_1(E_I + E_R)/n_2$. Substitute into tangential E:

$$(E_I - E_R)\cos\theta_i = \frac{n_1}{n_2}(E_I + E_R)\cos\theta_t$$

$$r_p = \frac{E_R}{E_I} = \frac{n_2\cos\theta_i - n_1\cos\theta_t}{n_2\cos\theta_i + n_1\cos\theta_t}$$

$$t_p = \frac{E_T}{E_I} = \frac{2n_1\cos\theta_i}{n_2\cos\theta_i + n_1\cos\theta_t}$$

Step 6: Verify normal incidence limit

At $\theta_i = \theta_t = 0$: $r_s = r_p = (n_1 - n_2)/(n_1 + n_2)$ and $t_s = t_p = 2n_1/(n_1 + n_2)$, recovering the normal-incidence result.

Step 1: Define reflectance and transmittance

Reflectance $R$ is the fraction of incident power reflected; transmittance $T$ is the fraction transmitted.

Step 2: Power flux normal to the interface

The time-averaged Poynting vector component normal to the interface for each wave:

$$\langle S_z \rangle_I = \frac{n_1}{2\mu_0 c}E_I^2\cos\theta_i, \quad \langle S_z \rangle_R = \frac{n_1}{2\mu_0 c}E_R^2\cos\theta_i, \quad \langle S_z \rangle_T = \frac{n_2}{2\mu_0 c}E_T^2\cos\theta_t$$

Step 3: Compute R

$$R = \frac{\langle S_z \rangle_R}{\langle S_z \rangle_I} = \frac{E_R^2}{E_I^2} = |r|^2$$

The reflectance is simply the square of the Fresnel reflection coefficient.

Step 4: Compute T

$$T = \frac{\langle S_z \rangle_T}{\langle S_z \rangle_I} = \frac{n_2\cos\theta_t}{n_1\cos\theta_i}\frac{E_T^2}{E_I^2} = \frac{n_2\cos\theta_t}{n_1\cos\theta_i}|t|^2$$

Note that $T \neq |t|^2$ in general due to the beam width change and impedance mismatch.

Step 5: Verify energy conservation R + T = 1

For s-polarization, substituting $r_s$ and $t_s$:

$$R_s + T_s = \left(\frac{n_1\cos\theta_i - n_2\cos\theta_t}{n_1\cos\theta_i + n_2\cos\theta_t}\right)^2 + \frac{n_2\cos\theta_t}{n_1\cos\theta_i}\left(\frac{2n_1\cos\theta_i}{n_1\cos\theta_i + n_2\cos\theta_t}\right)^2$$

Let $a = n_1\cos\theta_i$ and $b = n_2\cos\theta_t$:

$$= \frac{(a-b)^2}{(a+b)^2} + \frac{4ab}{(a+b)^2} = \frac{(a-b)^2 + 4ab}{(a+b)^2} = \frac{(a+b)^2}{(a+b)^2} = 1 \;\checkmark$$

Step 1: Set the p-polarization reflection coefficient to zero

$$r_p = \frac{n_2\cos\theta_i - n_1\cos\theta_t}{n_2\cos\theta_i + n_1\cos\theta_t} = 0$$

This requires the numerator to vanish:

$$n_2\cos\theta_B = n_1\cos\theta_t$$

Step 2: Use Snell's law to eliminate the transmitted angle

From Snell's law: $\sin\theta_t = (n_1/n_2)\sin\theta_B$, so:

$$\cos\theta_t = \sqrt{1 - \frac{n_1^2}{n_2^2}\sin^2\theta_B}$$

Step 3: Substitute and square both sides

$$n_2^2\cos^2\theta_B = n_1^2\cos^2\theta_t = n_1^2\left(1 - \frac{n_1^2}{n_2^2}\sin^2\theta_B\right)$$

$$n_2^2\cos^2\theta_B = n_1^2 - \frac{n_1^4}{n_2^2}\sin^2\theta_B$$

Step 4: Use $\cos^2\theta = 1 - \sin^2\theta$ and solve

$$n_2^2(1 - \sin^2\theta_B) = n_1^2 - \frac{n_1^4}{n_2^2}\sin^2\theta_B$$

$$n_2^2 - n_2^2\sin^2\theta_B = n_1^2 - \frac{n_1^4}{n_2^2}\sin^2\theta_B$$

$$\sin^2\theta_B\left(\frac{n_1^4}{n_2^2} - n_2^2\right) = n_1^2 - n_2^2$$

$$\sin^2\theta_B\cdot\frac{n_1^4 - n_2^4}{n_2^2} = n_1^2 - n_2^2$$

Step 5: Factor and simplify

Since $n_1^4 - n_2^4 = (n_1^2 - n_2^2)(n_1^2 + n_2^2)$:

$$\sin^2\theta_B = \frac{n_2^2}{n_1^2 + n_2^2}, \qquad \cos^2\theta_B = \frac{n_1^2}{n_1^2 + n_2^2}$$

$$\tan\theta_B = \frac{\sin\theta_B}{\cos\theta_B} = \frac{n_2}{n_1}$$

Step 6: Note that the reflected and transmitted rays are perpendicular

At Brewster's angle: $\theta_B + \theta_t = 90°$. Proof: $n_1\sin\theta_B = n_2\sin\theta_t = n_2\cos\theta_B$ gives $\tan\theta_B = n_2/n_1$ and $\sin\theta_t = \cos\theta_B = \sin(90°-\theta_B)$.

Physically, the reflected wave would need to oscillate along its own propagation direction, which is impossible for a transverse wave.

Step 1: Find the critical angle

From Snell's law: $\sin\theta_t = (n_1/n_2)\sin\theta_i$. Since $n_1 > n_2$, $\sin\theta_t > \sin\theta_i$.

When $\sin\theta_t = 1$ (maximum), we get the critical angle:

$$\sin\theta_c = \frac{n_2}{n_1}$$

Step 2: What happens beyond the critical angle?

For $\theta_i > \theta_c$: $\sin\theta_t = (n_1/n_2)\sin\theta_i > 1$. Then:

$$\cos\theta_t = \sqrt{1 - \sin^2\theta_t} = i\sqrt{\sin^2\theta_t - 1} \equiv i\beta$$

where $\beta = \sqrt{(n_1/n_2)^2\sin^2\theta_i - 1}$ is real and positive.

Step 3: The transmitted wave becomes evanescent

The transmitted wave's $z$-dependence: $e^{ik_{Tz}z}$ where $k_{Tz} = k_T\cos\theta_t = (n_2\omega/c)(i\beta)$:

$$\mathbf{E}_T \propto e^{ik_{Tx}x}\,e^{-\kappa z}\,e^{-i\omega t}$$

where $\kappa = (n_2\omega/c)\beta = (\omega/c)\sqrt{n_1^2\sin^2\theta_i - n_2^2}$.

Step 4: The field decays exponentially into medium 2

$$|\mathbf{E}_T| \propto e^{-\kappa z} = e^{-z/d}$$

The penetration depth is:

$$d = \frac{1}{\kappa} = \frac{c/\omega}{\sqrt{n_1^2\sin^2\theta_i - n_2^2}} = \frac{\lambda/(2\pi)}{\sqrt{n_1^2\sin^2\theta_i - n_2^2}}$$

Typically $d \sim \lambda$, so the evanescent wave extends about one wavelength into medium 2.

Step 5: Show R = 1 (total reflection)

Substituting $\cos\theta_t = i\beta$ into the Fresnel coefficient for s-polarization:

$$r_s = \frac{n_1\cos\theta_i - n_2(i\beta)}{n_1\cos\theta_i + n_2(i\beta)} = \frac{a - ib'}{a + ib'}$$

where $a = n_1\cos\theta_i$ and $b' = n_2\beta$ are both real. Therefore:

$$|r_s|^2 = \frac{a^2 + b'^2}{a^2 + b'^2} = 1 \implies R_s = 1$$

Similarly $R_p = 1$. All incident energy is reflected -- no time-averaged power flows into medium 2.

Step 6: Phase shift on total internal reflection

Although $|r| = 1$, the reflection coefficient acquires a phase:

$$r_s = e^{i\phi_s}, \quad \tan(\phi_s/2) = \frac{\sqrt{\sin^2\theta_i - (n_2/n_1)^2}}{\cos\theta_i}$$

This phase shift is exploited in the Fresnel rhomb to convert linear to circular polarization.