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Inside Collection (Course):

Course by: Paul Padley. E-mail the author

# The Fresnel Equations

Module by: Paul Padley. E-mail the author

Summary: We derive the Fresnel equations

## The Fresnel Equations

Snell's law, and the law of reflection are very useful for describing what happens to a ray of light at an interface. However we want to acheive a deeper understanding of what is happening to the electric field at the interface. We want to derive the relationships between the E E 's and B B 's. Hence we derive the Fresnel equations.

At an interface Any polarization of the fields can be broken down into simple components (which we will see later). Thus there are just two cases that we have to treat. Both of these cases are drawn at the scienceworld description of the Fresnel Equations http://scienceworld.wolfram.com/physics/FresnelEquations.html. We will first treat the case where the E E field is perpendicular to the plane of incidence. This corresponds to figure 20.1 in the book "Introduction to Optics" by Pedrotti and Pedrotti (second edition) and to figure 4.39 in the book "Optics" by Hecht (fourth edition). It is the second figure on the scienceworld web page. (It is important to note that while these three sources use similar conventions there are others, for example in this case we could have chosen to have the incoming B B field pointing upwards in the drawing. This would lead to different signs in the resulting equations.). We have in this case ( E E perpendicular): k ̂ × E = v B k ̂ × E = v B k ̂ E = 0 k ̂ E = 0 Also we know that E i + E r = E t E i + E r = E t (as was shown earlier) but since in this case the E E field is parallel to the interface we can write; E 0 i + E 0 r = E 0 t . E 0 i + E 0 r = E 0 t . Also we showed before that B i + B r μ i = B t μ t B i + B r μ i = B t μ t (where the is w.r.t. the surface). Note in the figure how the B B field changes direction and that the positive x direction is to the right in the figure. It is important to point out here, that we are evaluating things at the interface. At the interface the vector r r points to the plane of the interface. We have chosen a coordinate system so that the interface that lies at z = 0 z = 0 and contains the origin.

Then we rewrite the equation as B i μ i cos θ i B r μ i cos θ r = B t μ t cos θ t B i μ i cos θ i B r μ i cos θ r = B t μ t cos θ t You may ask, why cosine for a cross product? The cross product uses the complements of θ i θ i and θ t θ t which why it ends up being a cosine and not a sine. Using the law of reflection to subsitute for θ r θ r B i μ i cos θ i B r μ i cos θ i = B t μ t cos θ t B i μ i cos θ i B r μ i cos θ i = B t μ t cos θ t use B = E / v B = E / v E i v i μ i cos θ i E r v r μ i cos θ i = E t v t μ t cos θ t E i v i μ i cos θ i E r v r μ i cos θ i = E t v t μ t cos θ t use v i = v r v i = v r 1 μ i v i ( E i E r ) cos θ i = 1 μ t v t E t cos θ t 1 μ i v i ( E i E r ) cos θ i = 1 μ t v t E t cos θ t Now at the interface (using the arguments we have used before) we can write 1 μ i v i ( E 0 i E 0 r ) cos θ i = 1 μ t v t E 0 t cos θ t 1 μ i v i ( E 0 i E 0 r ) cos θ i = 1 μ t v t E 0 t cos θ t n i μ i ( E 0 i E 0 r ) cos θ i = n t μ t E 0 t cos θ t n i μ i ( E 0 i E 0 r ) cos θ i = n t μ t E 0 t cos θ t now we can substitute in E 0 i + E 0 r = E 0 t E 0 i + E 0 r = E 0 t n i μ i ( E 0 i E 0 r ) cos θ i = n t μ t ( E 0 i + E 0 r ) cos θ t n i μ i ( E 0 i E 0 r ) cos θ i = n t μ t ( E 0 i + E 0 r ) cos θ t Then rearrange or ( E 0 r E 0 i ) = n i μ i cos θ i n t μ t cos θ t n i μ i cos θ i + n t μ t cos θ t ( E 0 r E 0 i ) = n i μ i cos θ i n t μ t cos θ t n i μ i cos θ i + n t μ t cos θ t For the transmission we again start with n i μ i ( E 0 i E 0 r ) cos θ i = n t μ t E 0 t cos θ t n i μ i ( E 0 i E 0 r ) cos θ i = n t μ t E 0 t cos θ t and eliminate E 0 r E 0 r using E 0 i + E 0 r = E 0 t E 0 i + E 0 r = E 0 t or E 0 r = E 0 t E 0 i E 0 r = E 0 t E 0 i n i μ i ( 2 E 0 i E 0 t ) cos θ i = n t μ t E 0 t cos θ t n i μ i ( 2 E 0 i E 0 t ) cos θ i = n t μ t E 0 t cos θ t 2 n i μ i E 0 i cos θ i = ( n i μ i cos θ i + n t μ t cos θ t ) E 0 t 2 n i μ i E 0 i cos θ i = ( n i μ i cos θ i + n t μ t cos θ t ) E 0 t ( E 0 t E 0 i ) = 2 n i μ i cos θ i n i μ i cos θ i + n t μ t cos θ t ( E 0 t E 0 i ) = 2 n i μ i cos θ i n i μ i cos θ i + n t μ t cos θ t Now usually we only consider materials where μ i μ t μ 0 μ i μ t μ 0 so these equations simplify to: r ( E 0 r E 0 i ) = n i cos θ i n t cos θ t n i cos θ i + n t cos θ t r ( E 0 r E 0 i ) = n i cos θ i n t cos θ t n i cos θ i + n t cos θ t t ( E 0 t E 0 i ) = 2 n i cos θ i n i cos θ i + n t cos θ t t ( E 0 t E 0 i ) = 2 n i cos θ i n i cos θ i + n t cos θ t If the E E field is in the plane of incidence (the first figure on the scienceworld web page, figure 4.40 in Hecht, figure 20.2 in Pedrotti and Pedrotti) then the same sort of procedure can be followed (Pedrotti and Pedrott pages 409,410). It is left as a homework problem to show that in this case: r ( E 0 r E 0 i ) = n t cos θ i n i cos θ t n t cos θ i + n i cos θ t r ( E 0 r E 0 i ) = n t cos θ i n i cos θ t n t cos θ i + n i cos θ t t ( E 0 t E 0 i ) = 2 n i cos θ i n t cos θ i + n i cos θ t t ( E 0 t E 0 i ) = 2 n i cos θ i n t cos θ i + n i cos θ t

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