Fresnel Relations

The theory described in the previous section is sufficient to determine the directions of the reflected and refracted waves, when a light wave is obliquely incident on a plane interface between two dielectric media. However, it cannot determine the fractions of the incident energy that are reflected and refracted, respectively. In order to calculate the coefficients of reflection and transmission, it is necessary to take into account both the electric and the magnetic components of the various different waves. It turns out that there are two independent wave polarizations that behave slightly differently. The first of these is such that the magnetic components of the incident, reflected, and refracted waves are all parallel to the interface. The second is such that the electric components of these waves are all parallel to the interface.

Consider the first polarization. Let the interface correspond to the plane $z=0$ , let the region $z<0$ be occupied by material of refractive index $n_1$ , and let the region $z>0$ be occupied by material of refractive index $n_2$ . Suppose that the incident, reflected, and refracted waves are plane waves, of angular frequency $\omega$ , whose wavevectors lie in the $x$ -$z$ plane. (See Figure 14.) The equations governing the propagation of the wave are

$$\frac{\partial D_x}{\partial t} =-\frac{\partial H_y}{\partial z},$$ (956)

$$\frac{\partial D_z}{\partial t} =\frac{\partial H_y}{\partial x},$$ (957)

$$\frac{\partial H_y}{\partial t} =v^{\,2}\left(\frac{\partial D_z}{\partial x}- \frac{\partial D_x}{\partial z}\right),$$ (958)

where

$${\bf D} = \epsilon\,\epsilon_0\,{\bf E}$$ (959)

is the electric displacement, $v=c/n$ the characteristic wave speed, and $n=\sqrt{\epsilon}$ the refractive index. Suppose that, as described in the previous section,

$$\begin{multline} H_y(x,z,t) = \psi_i\,\cos(\omega\,t-n_1\,k_0\,\sin\theta_i\,x-n... ...os(\omega\,t-n_1\,k_0\,\sin\theta_i\,x+n_1\,k_0\,\cos\theta_i\,z) \end{multline}$$

in the region $z<0$ , and

$$H_y(x,z,t)= \psi_t\,\cos(\omega\,t-n_1\,k_0\,\sin\theta_i\,x-n_2\,k_0\,\cos\theta_t\,z)$$ (960)

in the region $z>0$ . Here, $k_0=\omega/c$ is the vacuum wavenumber, $\theta_i$ the angle of incidence, and $\theta_t$ the angle of refraction. (See Figure 14.) In writing the above expressions, we have made use of the law of reflection (i.e., $\theta_r=\theta_i$ ), as well as the law of refraction (i.e., $n_1\,\sin\theta_i=n_2\,\sin\theta_t$ ). The two terms on the right-hand side of Equation (961) correspond to the incident and reflected waves, respectively. The term on the right-hand side of Equation (962) corresponds to the refracted wave. Substitution of Equations (961)-(962) into the governing differential equations, (957)-(959), yields

$$D_x(x,z,t) = \frac{\psi_i\,\cos\theta_i}{v_1}\,\cos(\omega\,t-n_1\,k_0\,\sin\theta_i\,x-n_1\,k_0\,\cos\theta_i\,z)$$

$$~~~~-\frac{\psi_r\,\cos\theta_i}{v_1}\,\cos(\omega\,t-n_1\,k_0\,\sin\theta_i\,x+n_1\,k_0\,\cos\theta_i\,z),$$ (961)

$$D_z(x,z,t) = -\frac{\psi_i\,\sin\theta_i}{v_1}\,\cos(\omega\,t-n_1\,k_0\,\sin\theta_i\,x-n_1\,k_0\,\cos\theta_i\,z)$$

$$~~~~-\frac{\psi_r\,\sin\theta_i}{v_1}\,\cos(\omega\,t-n_1\,k_0\,\sin\theta_i\,x+n_1\,k_0\,\cos\theta_i\,z)$$ (962)

in the region $z<0$ , and

$$D_x(x,z,t) =\frac{\psi_t\,\cos\theta_t}{v_2}\,\cos(\omega\,t-n_1\,k_0\,\sin\theta_i\,x-n_2\,k_0\,\cos\theta_t\,z),$$ (963)

$$D_z(x,z,t) =-\frac{\psi_t\,\sin\theta_i}{v_1}\,\cos(\omega\,t-n_1\,k_0\,\sin\theta_i\,x-n_2\,k_0\,\cos\theta_t\,z)$$ (964)

in the region $z>0$ .

Now, both the normal and the tangential components of the magnetic intensity must be continuous at the interface. This implies that

$$[H_y]_{z=0_-}^{z=0_+}=0,$$ (965)

which yields

$$\psi_i+\psi_r=\psi_t$$ (966)

Furthermore, the normal component of the electric displacement, as well as the tangential component of the electric field, must be continuous at the interface. In other words,

$$[D_z]_{z=0_-}^{z=0^+} = 0,$$ (967)

and

$$[E_x]_{z=0_-}^{z=0_+}=[D_x/(\epsilon\,\epsilon_0)]_{z=0_-}^{z=0_+} = 0.$$ (968)

The former of these conditions again gives Equation (968), whereas the latter yields

$$\psi_i-\psi_r= \frac{\alpha}{\beta}\,\psi_t.$$ (969)

Here,

$$\alpha =\frac{\cos\theta_t}{\cos\theta_i},$$ (970)

$$\beta =\frac{v_1}{v_2}=\frac{n_2}{n_1}.$$ (971)

It follows that

$$\psi_r =\left(\frac{\beta-\alpha}{\beta+\alpha}\right)\psi_i,$$ (972)

$$\psi_t =\left(\frac{2\,\beta}{\beta+\alpha}\right)\psi_i.$$ (973)

The electromagnetic energy flux in the $z$ -direction (i.e., normal to the interface) is

$${\cal I}_z = E_x\,H_y.$$ (974)

Thus, the mean energy fluxes associated with the incident, reflected, and refracted waves are

$$\langle {\cal I}_z\rangle_i = \frac{\psi_i^{\,2}\,\cos\theta_i}{2\,\epsilon_0\,c\,n_1},$$ (975)

$$\langle {\cal I}_z\rangle_r = -\frac{\psi_r^{\,2}\,\cos\theta_i}{2\,\epsilon_0\,c\,n_1},$$ (976)

$$\langle {\cal I}_z\rangle_t = \frac{\psi_t^{\,2}\,\cos\theta_t}{2\,\epsilon_0\,c\,n_2},$$ (977)

respectively. The coefficients of reflection and transmission are defined

$$R = \frac{-\langle {\cal I}_z\rangle_r}{\langle {\cal I}_z\rangle_i},$$ (978)

$$T = \frac{\langle {\cal I}_z\rangle_t}{\langle {\cal I}_z\rangle_i},$$ (979)

respectively. Hence, it follows that

$$R =\left(\frac{\beta-\alpha}{\beta+\alpha}\right)^2,$$ (980)

$$T =\frac{4\,\alpha\,\beta}{(\beta+\alpha)^2} = 1-R.$$ (981)

These expressions are known as Fresnel relations.

Let us now consider the second polarization, in which the electric components of the incident, reflected, and refracted waves are all parallel to the interface. In this case, the governing equations are

$$\frac{\partial H_x}{\partial t} =v^{\,2}\,\frac{\partial D_y}{\partial z},$$ (982)

$$\frac{\partial H_z}{\partial t} =-v^{\,2}\,\frac{\partial D_y}{\partial x},$$ (983)

$$\frac{\partial D_y}{\partial t} =-\frac{\partial H_z}{\partial x}+ \frac{\partial H_x}{\partial z}.$$ (984)

If we make the transformations $H_y\rightarrow -v^{\,2}\,D_y$ , $D_x\rightarrow H_x$ , $D_z\rightarrow H_z$ , $\psi_{i,r,t}\rightarrow -v\,\psi_{i,r,t}$ then we can reuse the solutions that we derived for the other polarization. We find that

$$D_y(x,z,t) = \frac{\psi_i}{v_1}\,\cos(\omega\,t-n_1\,k_0\,\sin\theta_i\,x-n_1\,k_0\,\cos\theta_i\,z)$$

$$~~~~+\frac{\psi_r}{v_1}\,\cos(\omega\,t-n_1\,k_0\,\sin\theta_i\,x+n_1\,k_0\,\cos\theta_i\,z),$$ (985)

$$H_x(x,z,t) =-\psi_i\,\cos\theta_i\,\cos(\omega\,t-n_1\,k_0\,\sin\theta_i\,x-n_1\,k_0\,\cos\theta_i\,z)$$

$$~~~~+\psi_r\,\cos\theta_i\,\cos(\omega\,t-n_1\,k_0\,\sin\theta_i\,x+n_1\,k_0\,\cos\theta_i\,z),$$ (986)

$$H_z(x,z,t) = \psi_i\,\sin\theta_i\,\cos(\omega\,t-n_1\,k_0\,\sin\theta_i\,x-n_1\,k_0\,\cos\theta_i\,z)$$

$$~~~~+\psi_r\,\sin\theta_i\,\cos(\omega\,t-n_1\,k_0\,\sin\theta_i\,x+n_1\,k_0\,\cos\theta_i\,z)$$ (987)

in the region $z<0$ , and

$$D_y(x,z,t) = \frac{\psi_t}{v_2}\,\cos(\omega\,t-n_1\,k_0\,\sin\theta_i\,x-n_2\,k_0\,\cos\theta_t\,z),$$ (988)

$$H_x(x,z,t) =-\psi_t\,\cos\theta_t\,\cos(\omega\,t-n_1\,k_0\,\sin\theta_i\,x-n_2\,k_0\,\cos\theta_t\,z),$$ (989)

$$H_z(x,z,t) =\psi_t\,\sin\theta_t\,\cos(\omega\,t-n_1\,k_0\,\sin\theta_i\,x-n_2\,k_0\,\cos\theta_t\,z)$$ (990)

in the region $z>0$ . The first two matching conditions at the interface are that the normal and tangential components of the magnetic intensity are continuous. In other words,

$$[H_z]_{z=0_+}^{z=0_-} =0,$$ (991)

$$[H_x]_{z=0_+}^{z=0_-} =0.$$ (992)

The first of these conditions yields

$$\psi_i+\psi_r = \beta^{\,-1}\,\psi_t,$$ (993)

whereas the second gives

$$\psi_i-\psi_r = \alpha\,\psi_t.$$ (994)

The final matching condition at the interface is that the tangential component of the electric field is continuous. In other words,

$$[E_y]_{z=0_-}^{z=0_+} = [D_y/(\epsilon\,\epsilon_0)]_{z=0_-}^{z=0_+}=0,$$ (995)

which again yields Equation (995). It follows that

$$\psi_r =\left(\frac{1-\alpha\,\beta}{1+\alpha\,\beta}\right)\psi_i,$$ (996)

$$\psi_t =\left(\frac{2\,\beta}{1+\alpha\,\beta}\right)\psi_i.$$ (997)

The electromagnetic energy flux in the $z$ -direction is

$${\cal I}_z = -E_y\,H_x.$$ (998)

Thus, the mean energy fluxes associated with the incident, reflected, and refracted waves are

$$\langle {\cal I}_z\rangle_i = \frac{\psi_i^{\,2}\,\cos\theta_i}{2\,\epsilon_0\,c\,n_1},$$ (999)

$$\langle {\cal I}_z\rangle_r = -\frac{\psi_r^{\,2}\,\cos\theta_i}{2\,\epsilon_0\,c\,n_1},$$ (1000)

$$\langle {\cal I}_z\rangle_t = \frac{\psi_t^{\,2}\,\cos\theta_t}{2\,\epsilon_0\,c\,n_2},$$ (1001)

respectively. Hence, the coefficients of reflection and transmission are

$$R =\left(\frac{1-\alpha\,\beta}{1+\alpha\,\beta}\right)^2,$$ (1002)

$$T =\frac{4\,\alpha\,\beta}{(1+\alpha\,\beta)^2} = 1-R,$$ (1003)

respectively. These expressions are the Fresnel relations for the polarization in which the electric field is parallel to the interface.

Figure: 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. The Brewster angle is $56.3^\circ$ .

It can be seen that, at oblique incidence, the Fresnel relations (982) and (983) for the polarization in which the magnetic field is parallel to the interface are different to the corresponding relations (1004) and (1005) for the polarization in which the electric field is parallel to the interface. This implies that the coefficients of reflection and transmission for these two polarizations are, in general, different.

Figure 15 shows the coefficients of reflection and transmission for oblique incidence from air ($n_1=1.0$ ) to glass ($n_2=1.5$ ). Roughly speaking, it can be seen that the coefficient of reflection rises, and the coefficient of transmission falls, as the angle of incidence increases. However, for the polarization in which the magnetic field is parallel to the interface, there is a particular angle of incidence, know as the Brewster angle, at which the reflected intensity is zero. There is no similar behavior for the polarization in which the electric field is parallel to the interface.

It follows from Equation (982) that the Brewster angle corresponds to the condition

$$\alpha=\beta,$$ (1004)

or

$$\beta^{\,2} = \frac{\cos^2\theta_t}{\cos^2\theta_i} = \frac{1-\si... ...eta_t}{1-\sin^2\theta_i}=\frac{1-\sin^2\theta_i/\beta^{\,2}}{1-\sin^2\theta_i},$$ (1005)

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

$$\sin\theta_i = \frac{\beta}{\sqrt{1+\beta^{\,2}}},$$ (1006)

or $\tan\theta_i = \beta = n_2/n_1$ . Hence, the Brewster angle corresponds to $\theta_i=\theta_B$ , where

$$\theta_B = \tan^{-1}\left(\frac{n_2}{n_1}\right).$$ (1007)

If unpolarized light is incident on an air/glass (say) interface at the Brewster angle then the reflected light is $100$ percent linearly polarized. (See Section 7.6.)

The fact that the coefficient of reflection for the polarization in which the electric field is parallel to the interface is generally greater that that for the other polarization (see Figure 15) implies that sunlight reflected from a horizontal water or snow surface is partially linearly polarized, with the horizontal polarization predominating over the vertical one. Such reflected light may be so intense as to cause glare. Polaroid sunglasses help reduce this glare by blocking horizontally polarized light.