Fresnel Relations

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

where

(959) |

is the electric displacement, the characteristic wave speed, and the refractive index. Suppose that, as described in the previous section,

in the region , and

in the region . Here, is the vacuum wavenumber, the angle of incidence, and the angle of refraction. (See Figure 14.) In writing the above expressions, we have made use of the law of reflection (i.e., ), as well as the law of refraction (i.e., ). 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

(961) | ||

(962) |

in the region , and

(963) | ||

(964) |

in the region .

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

which yields

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,

and

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

(969) |

Here,

(970) | ||

(971) |

It follows that

(972) | ||

(973) |

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

(974) |

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

(975) | ||

(976) | ||

(977) |

respectively. The coefficients of reflection and transmission are defined

(978) | ||

(979) |

respectively. Hence, it follows that

These expressions are known as

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

(982) | ||

(983) | ||

(984) |

If we make the transformations , , , then we can reuse the solutions that we derived for the other polarization. We find that

(985) | ||

(986) | ||

(987) |

in the region , and

(988) | ||

(989) | ||

(990) |

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

The first of these conditions yields

whereas the second gives

(994) |

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

which again yields Equation (995). It follows that

(996) | ||

(997) |

The electromagnetic energy flux in the -direction is

(998) |

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

(999) | ||

(1000) | ||

(1001) |

respectively. Hence, the coefficients of reflection and transmission are

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

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 (
) to
glass (
). 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

(1004) |

or

(1005) |

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

(1006) |

or . Hence, the Brewster angle corresponds to , where

(1007) |

If unpolarized light is incident on an air/glass (say) interface at the Brewster angle then the reflected light is 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.