Reflection at a dielectric boundary

An electromagnetic wave of real (positive) frequency $\omega$ can be written

$${\bf E}({\bf r}, t) = {\bf E}_0 {\rm e}^{ {\rm i} ({\bf k}\cdot{\bf r} - \omega t)},$$ (1210)

$${\bf B}({\bf r}, t) = {\bf B}_0 {\rm e}^{ {\rm i} ({\bf k}\cdot{\bf r}-\omega t)}.$$ (1211)

The wave-vector, ${\bf k}$, indicates the direction of propagation of the wave, and also its phase-velocity, $v$, via

$$v = \frac{\omega}{k}.$$ (1212)

Since the wave is transverse in nature, we must have ${\bf E}_0\cdot{\bf k} = {\bf B}_0\cdot{\bf k} = 0$. Finally, the familiar Maxwell equation

$$\nabla\times {\bf E} = -\frac{\partial {\bf B}}{\partial t}$$ (1213)

leads us to the following relation between the constant vectors ${\bf E}_0$ and ${\bf B}_0$:

$${\bf B}_0 = \frac{\hat{\bf k}\times {\bf E}_0}{v}.$$ (1214)

Here, $\hat{\bf k} = {\bf k}/k$ is a unit vector pointing in the direction of wave propagation.

Suppose that the plane $z=0$ forms the boundary between two different dielectric media. Let medium 1, of refractive index $n_1$, occupy the region $z<0$, whilst medium 2, of refractive index $n_2$, occupies the region $z>0$. Let us investigate what happens when an electromagnetic wave is incident on this boundary from medium 1.

Figure 55:

Consider, first of all, the simple case of incidence normal to the boundary (see Fig. 55). In this case, $\hat{\bf k} = +\hat{\bf z}$ for the incident and transmitted waves, and $\hat{\bf k} = -\hat{\bf z}$ for the reflected wave. Without loss of generality, we can assume that the incident wave is polarized in the $x$-direction. Hence, using Eq. (1214), the incident wave can be written

$${\bf E}(z,t) = E_i {\rm e}^{ {\rm i} (k_1 z-\omega t)} \hat{\bf x},$$ (1215)

$${\bf B}(z,t) = \frac{E_i}{v_1} {\rm e}^{ {\rm i} (k_1 z-\omega t)} \hat{\bf y},$$ (1216)

where $v_1=c/n_1$ is the phase-velocity in medium 1, and $k_1=\omega/v_1$. Likewise, the reflected wave takes the form

$${\bf E}(z,t) = E_r {\rm e}^{ {\rm i} (-k_1 z-\omega t)} \hat{\bf x},$$ (1217)

$${\bf B}(z,t) = -\frac{E_r}{v_1} {\rm e}^{ {\rm i} (-k_1 z-\omega t)} \hat{\bf y}.$$ (1218)

Finally, the transmitted wave can be written

$${\bf E}(z,t) = E_t {\rm e}^{ {\rm i} (k_2 z-\omega t)} \hat{\bf x},$$ (1219)

$${\bf B}(z,t) = \frac{E_t}{v_2} {\rm e}^{ {\rm i} (k_2 z-\omega t)} \hat{\bf y},$$ (1220)

where $v_2=c/n_2$ is the phase-velocity in medium 2, and $k_2=\omega/v_2$.

For the case of normal incidence, the electric and magnetic components of all three waves are parallel to the boundary between the two dielectric media. Hence, the appropriate boundary conditions to apply at $z=0$ are

$$E_{\parallel 1} = E_{\parallel 2},$$ (1221)

$$B_{\parallel 1} = B_{\parallel 2}.$$ (1222)

The latter condition derives from the general boundary condition $H_{\parallel 1} = H_{\parallel 2}$, and the fact that ${\bf B}=\mu_0 {\bf H}$ in both media (which are assumed to be non-magnetic).

Application of the boundary condition $(\ref{e9.147})$ yields

$$E_i + E_r = E_t.$$ (1223)

Likewise, application of the boundary condition (1222) gives

$$\frac{E_i - E_r}{v_1} = \frac{E_t}{v_2},$$ (1224)

or

$$E_i - E_r = \frac{v_1}{v_2} E_t = \frac{n_2}{n_1} E_t,$$ (1225)

since $v_1/v_2=n_2/n_1$. Equations (1223) and (1225) can be solved to give

$$E_r = \left(\frac{n_1-n_2}{n_1+n_2}\right) E_i,$$ (1226)

$$E_t = \left(\frac{2 n_1}{n_1+n_2}\right) E_t.$$ (1227)

Thus, we have determined the amplitudes of the reflected and transmitted waves in terms of the amplitude of the incident wave.

It can be seen, first of all, that if $n_1=n_2$ then $E_r=0$ and $E_t=E_i$. In other words, if the two media have the same indices of refraction then there is no reflection at the boundary between them, and the transmitted wave is consequently equal in amplitude to the incident wave. On the other hand, if $n_1\neq n_2$ then there is some reflection at the boundary. Indeed, the amplitude of the reflected wave is roughly proportional to the difference between $n_1$ and $n_2$. This has important practical consequences. We can only see a clean pane of glass in a window because some of the light incident at an air/glass boundary is reflected, due to the different refractive indicies of air and glass. As is well-known, it is a lot more difficult to see glass when it is submerged in water. This is because the refractive indices of glass and water are quite similar, and so there is very little reflection of light incident on a water/glass boundary.

According to Eq. (1226), $E_r/E_i<0$ when $n_2> n_1$. The negative sign indicates a $180^\circ$ phase-shift of the reflected wave, with respect to the incident wave. We conclude that there is a $180^\circ$ phase-shift of the reflected wave, relative to the incident wave, on reflection from a boundary with a medium of greater refractive index. Conversely, there is no phase-shift on reflection from a boundary with a medium of lesser refractive index.

The mean electromagnetic energy flux, or intensity, in the $z$-direction is simply

$$I =\frac{\langle {\bf E}\times{\bf B}\cdot\hat{\bf z}\rangle... ...0}= \frac{E_0 B_0}{2 \mu_0} = \frac{E_0^{ 2}}{2 \mu_0 v}.$$ (1228)

The coefficient of reflection, $R$, is defined as the ratio of the intensities of the reflected and incident waves:

$$R = \frac{I_r}{I_i} = \left(\frac{E_r}{E_i}\right)^2.$$ (1229)

Likewise, the coefficient of transmission, $T$, is the ratio of the intensities of the transmitted and incident waves:

$$T = \frac{I_t}{I_i} =\frac{v_1}{v_2}\left(\frac{E_t}{E_i}\right)^2=\frac{n_2}{n_1}\left(\frac{E_t}{E_i}\right)^2.$$ (1230)

Equations (1226), (1227), (1229), and (1230) yield

$$R = \left(\frac{n_1-n_2}{n_1+n_2}\right)^2,$$ (1231)

$$T = \frac{n_2}{n_1}\left(\frac{2 n_1}{n_1+n_2}\right)^2.$$ (1232)

Note that $R+T=1$. In other words, any wave energy which is not reflected at the boundary is transmitted, and vice versa.

Figure 56:

Let us now consider the case of incidence oblique to the boundary (see Fig. 56). Suppose that the incident wave subtends an angle $\theta_i$ with the normal to the boundary, whereas the reflected and transmitted waves subtend angles $\theta_r$ and $\theta_t$, respectively.

The incident wave can be written

$${\bf E}({\bf r}, t) = {\bf E}_i {\rm e}^{ {\rm i} ({\bf k}_i\cdot{\bf r}-\omega t)},$$ (1233)

$${\bf B}({\bf r}, t) = {\bf B}_i {\rm e}^{ {\rm i} ({\bf k}_i\cdot{\bf r}-\omega t)},$$ (1234)

with analogous expressions for the reflected and transmitted waves. Since, in the case of oblique incidence, the electric and magnetic components of the wave fields are no longer necessarily parallel to the boundary, the boundary conditions (1221) and (1222) at $z=0$ must be supplemented by the additional boundary conditions

$$\epsilon_1 E_{\perp 1} = \epsilon_2 E_{\perp 2},$$ (1235)

$$B_{\perp 1} = B_{\perp 2}.$$ (1236)

Equation (1235) derives from the general boundary condition $D_{\perp 1} = D_{\perp 2}$.

It follows from Eqs. (1222) and (1236) that both components of the magnetic field are continuous at the boundary. Hence, we can write

$${\bf B}_i {\rm e}^{ {\rm i} ({\bf k}_i\cdot{\bf r}-\omeg... ...f B}_t {\rm e}^{ {\rm i} ({\bf k}_t\cdot{\bf r}-\omega t)}$$ (1237)

at $z=0$. Given that ${\bf B}_i$, ${\bf B}_r$, and ${\bf B}_t$ are constant vectors, the only way in which the above equation can be satisfied for all values of $x$ and $y$ is if

$${\bf k}_i\cdot{\bf r} = {\bf k}_r\cdot{\bf r} = {\bf k}_t\cdot{\bf r}$$ (1238)

throughout the $z=0$ plane. This, in turn, implies that

$$k_{i x} =k_{r x}= k_{t x}$$ (1239)

and

$$k_{i y} = k_{r y} = k_{t y}.$$ (1240)

It immediately follows that if $k_{i y}=0$ then $k_{r y} =k_{t y}=0$. In other words, if the incident wave lies in the $x$-$z$ plane then the reflected and transmitted waves also lie in the $x$-$z$ plane. Another way of putting this is that the incident, reflected, and transmitted waves all lie in the same plane, know as the plane of incidence. This, of course, is one of the laws of geometric optics. From now on, we shall assume that the plane of incidence is the $x$-$z$ plane.

Now, $k_i=k_r = \omega/v_1$ and $k_t=\omega/v_2$. Moreover,

$$\sin\theta_i = \frac{k_{x i}}{k_i},$$ (1241)

with similar expressions for $\theta_r$ and $\theta_t$. Hence, according to Eq. (1239),

$$\sin\theta_r = \sin\theta_i,$$ (1242)

which implies that $\theta_r=\theta_i$. Moreover,

$$\frac{\sin\theta_t}{\sin\theta_i} = \frac{v_2}{v_1} = \frac{n_1}{n_2}.$$ (1243)

Of course, the above expressions correspond to the law of reflection and Snell's law of refraction, respectively.

For the case of oblique incidence, we need to consider two independent wave polarizations separately. The first polarization has all the wave electric fields perpendicular to the plane of incidence, whilst the second has all the wave magnetic fields perpendicular to the plane of incidence.

Let us consider the first wave polarization. We can write unit vectors in the directions of propagation of the incident, reflected, and transmitted waves likso:

$$\hat{\bf k}_i = \left(\sin\theta_i, 0, \cos\theta_i\right),$$ (1244)

$$\hat{\bf k}_r = \left(\sin\theta_i, 0, -\cos\theta_i\right),$$ (1245)

$$\hat{\bf k}_t = \left(\sin\theta_t, 0, \cos\theta_t\right).$$ (1246)

The constant vectors associated with the incident wave are written

$${\bf E}_i = E_i \hat{\bf y},$$ (1247)

$${\bf B}_i = \frac{E_i}{v_1} \left(-\cos\theta_i, 0, \sin\theta_i\right),$$ (1248)

where use has been made of Eq. (1214). Likewise, the constant vectors associated with the reflected and transmitted waves are

$${\bf E}_r = E_r \hat{\bf y},$$ (1249)

$${\bf B}_r = \frac{E_r}{v_1} \left(\cos\theta_i, 0, \sin\theta_i\right),$$ (1250)

and

$${\bf E}_t = E_t \hat{\bf y},$$ (1251)

$${\bf B}_t = \frac{E_t}{v_2} \left(-\cos\theta_t, 0, \sin\theta_t\right),$$ (1252)

respectively.

Now, the boundary condition (1221) yields $E_{y 1} = E_{y 2}$, or

$$E_i + E_r = E_t.$$ (1253)

Likewise, the boundary condition (1236) gives $B_{z 1} = B_{z 2}$, or

$$(E_i + E_r) \frac{\sin\theta_i}{v_1} = E_t \frac{\sin\theta_t}{v_2}.$$ (1254)

However, using Snell's law, (1243), the above expression reduces to Eq. (1253). Finally, the boundary condition (1222) yields $B_{x 1} = B_{x 2}$, or

$$(E_i -E_r) \frac{\cos\theta_i}{v_1} = E_t \frac{\cos\theta_t}{v_2}.$$ (1255)

It is convenient to define the parameters

$$\alpha =\frac{\cos\theta_t}{\cos\theta_i},$$ (1256)

and

$$\beta = \frac{v_1}{v_2}= \frac{n_2}{n_1}.$$ (1257)

Equations (1253) and (1255) can be solved in terms of these parameters to give

$$E_r = \left(\frac{1-\alpha \beta}{1+\alpha \beta}\right) E_i,$$ (1258)

$$E_t = \left(\frac{2}{1+\alpha \beta}\right) E_i.$$ (1259)

These relations are known as Fresnel equations.

The wave intensity in the $z$-direction is given by

$$I_z =\frac{\langle {\bf E}\times{\bf B}\cdot\hat{\bf z}\rang... ...\theta}{2 \mu_0} = \frac{E_0^{ 2} \cos\theta}{2 \mu_0 v}.$$ (1260)

Hence, the coefficient of reflection is written

$$R = \left(\frac{E_r}{E_i}\right)^2 = \left(\frac{1-\alpha \beta}{1+\alpha \beta}\right)^2,$$ (1261)

whereas the coefficient of transmission takes the form

$$T = \frac{\cos\theta_t}{\cos\theta_i} \frac{v_1}{v_2}\left(... ...t)^2 = \alpha \beta \left(\frac{2}{1+\alpha \beta}\right)^2.$$ (1262)

Note that it is again the case that $R+T=1$.

Let us now consider the second wave polarization. In this case, the constant vectors associated with the incident, reflected, and transmitted waves are written

$${\bf E}_i = E_i (\cos\theta_i, 0, -\sin\theta_i),$$ (1263)

$${\bf B}_i = \frac{E_i}{v_1} \hat{\bf y},$$ (1264)

and

$${\bf E}_r = E_r (\cos\theta_i, 0, \sin\theta_i),$$ (1265)

$${\bf B}_r = - \frac{E_r}{v_1} \hat{\bf y},$$ (1266)

and

$${\bf E}_t = E_t (\cos\theta_t, 0, -\sin\theta_t),$$ (1267)

$${\bf B}_t = \frac{E_t}{v_2} \hat{\bf y},$$ (1268)

respectively. The boundary condition (1222) yields $B_{y 1}= B_{y 2}$, or

$$\frac{E_i-E_r}{v_1} = \frac{E_t}{v_2}.$$ (1269)

Likewise, the boundary condition (1221) gives $E_{x 1}=E_{x 2}$, or

$$(E_i+E_r) \cos\theta_i = E_t \cos\theta_t.$$ (1270)

Finally, the boundary condition (1235) yields $\epsilon_1 E_{z 1} = \epsilon_2 E_{z 2}$, or

$$\epsilon_1 (E_i-E_r) \sin\theta_i = \epsilon_2 E_i \sin\theta_t.$$ (1271)

Making use of Snell's law, and the fact that $\epsilon = n^2$, the above expression reduces to Eq. (1269).

Solving Eqs. (1239) and (1270), we obtain

$$E_r = \left(\frac{\alpha-\beta}{\alpha+\beta}\right) E_i,$$ (1272)

$$E_t = \left(\frac{2}{\alpha+\beta}\right) E_i.$$ (1273)

The associated coefficients of reflection and transmission take the form

$$R = \left(\frac{\alpha-\beta}{\alpha+\beta}\right)^2,$$ (1274)

$$T = \alpha \beta\left(\frac{2}{\alpha+\beta}\right)^2,$$ (1275)

respectively. As usual, $R+T=1$.

Note that at oblique incidence the Fresnel equations, (1258) and (1259), for the wave polarization in which the electric field is parallel to the boundary are different to the Fresnel equations, (1272) and (1273), for the wave polarization in which the magnetic field is parallel to the boundary. This implies that the coefficients of reflection and transmission for these two wave polarizations are, in general, different.

Figure 57:

Figure 57 shows the coefficients of reflection (solid curves) and transmission (dashed curves) for oblique incidence from air ($n_1=1.0$) to glass ($n_2=1.5$). The left-hand panel shows the wave polarization for which the electric field is parallel to the boundary, whereas the right-hand panel shows the wave polarization for which the magnetic field is parallel to the boundary. In general, it can be seen that the coefficient of reflection rises, and the coefficient of transmission falls, as the angle of incidence increases. Note, however, that for the second wave polarization there is a particular angle of incidence, know as the Brewster angle, at which the reflected intensity is zero. There is no similar behaviour for the first wave polarization.

It follows from Eq. (1272) that the Brewster angle corresponds to the condition

$$\alpha=\beta,$$ (1276)

or

$$\beta^2 = \frac{\cos^2\theta_t}{\cos^2\theta_i} = \frac{1-\s... ...^2\theta_i}=\frac{1-\sin^2\theta_i/\beta^2}{1-\sin^2\theta_i},$$ (1277)

where use has been made of Snell's law. The above expression reduces to

$$\sin\theta_i = \frac{\beta}{\sqrt{1+\beta^2}},$$ (1278)

or $\tan\theta_i = \beta = n_2/n_1$. Hence, the Brewster angle satisfies

$$\theta_B = \tan^{-1}\left(\frac{n_2}{n_1}\right).$$ (1279)

If unpolarized light is incident on an air/glass (say) boundary at the Brewster angle then the reflected beam is $100\%$ plane polarized.