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The Fresnel Equations
for Reflectance and Transmission


Augustin Jean Fresnel

In the early part of the 19th century when the laws of optics were being worked out Augustin Jean Fresnel moved the development forward by a major amount with his equations which gave the degree of reflectance and transmission of a beam of light impinging upon the interface between two media. The material here presents these equations, shows their alternate form, their use and finally how they were derived.

First there is the matter of the pronunciation of Fresnel. It is French and therefore any final consonant in a syllable except c,r,f and l (those in careful) are not pronounced. Therefore Fresnel is properly pronounced as fray nell.

Below is a diagram showing the relevant quantities for a beam of light, called the incident beam, impinging upon an interface between two media of possibly different indices of refraction, ni and nt. It is notable that the incident beam, the reflected beam and the transmitted beam are all in the same plane, called the incidence plane. The subscripts i, r and t refer to incident, reflected and transmitted, respectively.

The blue line in the above diagram represents a line perpendicular to the interface surface, a normal. The angles θi, θr, and θt are measured from the normal.

There were two results that preceded Fresnel's analysis:

θi = θr
and Snell's Law
nisin(θi) = ntsin(θt)

The Equations of Fresnel

Fresnel was working many decades before James Clerk Maxwell formulated his theory of electro-magnetism that explained radiation, including light as electromagnetic wave oscillation. In Fresnel's time optics was just discovering the notion of light as wave phenomena. The field had been dominated by the corpuscular theory of light which Isaac Newton propounded. Working under these conceptual limitations it is quite remarkable how accurate Fresnel's analysis was. Fresnel believed that the radiation was actual vibration in an all-pervading material aether. In the material below use will be made of the concepts which came after Fresnel.

Let Ei stand for the amplitude of the wave oscillation in the incident beam. Er and Et are the amplitudes for the reflected and transmitted beams. The beams can be decomposed into a component which is perpendicular to the incidence plane and one that is parallel to the incidence plane.

The amplitude coefficients of reflection and transmission, r and t, are defined as

r = Er/Ei
t = Et/Ei

Fresnel's results stem from the requirement that the amplitude of the components of the beams parallel to the interfact surface must be continuous.

The Fresnel equations are then for the component perpendicular to the incidence plane:

r = [nicos(θi)− ntcos(θt)]/[nicos(θi) + ntcos(θt)]
t = [2nicos(θi)]/[nicos(θi) + ntcos(θt)]

For the component parallel to the incidence plane they are:

r = [nicos(θi)− ntcos(θi)]/[nicos(θt) + ntcos(θi)]
t = [2nicos(θi)]/[nicos(θt) + ntcos(θi)]

These equations are a bit cumbersome. More elegant versions can be derived; i.e.,

perpendicular r = −[sin(θi)−sin(θt)]/[sin(θi)+sin(θt)]
and
parallel r =[tan(θi)−tan(θt)]/[tan(θi)+tan(θt)]

Power transmission and reflectance

The power of electromagnetic radiation is proportional to product of the electric field ampliture and the magnetic field amplitude, EB, but the magnet field amplitude is proportional to the electric field. Therefore the proportion of incident power reflected is equal to r². This is called the reflectance R. The transmittance T, the proportion of power transmitted is not equal to t², instead to (nt/ni)(cos(θt)/cos(θi))t². It is usually easier to get the transmittance T from the relation

T = 1 − R = 1 − r² Thus t² = 1 − r².

Consider the case of a light mean passing from air (index of refraction ni=1) into diamond (index of refraction nt=2.4). Suppose the angle of the incident beam is 45° with respect to the surgace normal. The angle of reflection is then 45°. By Snell's Law the angle of transmission is given by:

sin((θi) = (1/2.4)sin(45°) = 0.2946
and thus
θi = 17.135°

Therefore r is given by

perpendicular: r = −[0.7071 − 0.2946]/[0.7071 + 0.2946] = −0.4118
parallel: r = [1.0000 − 0.3083]/[1.0000 +0.3083] = 0.5287
r² = (−0.4118)² + (0.5287)² = 0.4491.

Thus 44.91 percent of the incident beam is reflected and therefore 55.1 percent is transmitted.

(To be continued.)


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