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Phase Changes

Module by: Paul Padley. E-mail the author

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Summary: We look at what happens to the phase of a wave upon reflection.

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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

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 We can rewrite these equations using Snell's Law to eliminate the cos θ t cos θ t term. From simple trigonometry we know that cos θ t = 1 − sin 2 θ t . cos θ t = 1 − sin 2 θ t . We also know from Snell's law that sin θ t = n i n t sin θ i sin θ t = n i n t sin θ i so we have cos θ t = 1 − n i 2 n t 2 sin 2 θ i . cos θ t = 1 − n i 2 n t 2 sin 2 θ i .

We can substitute this into r ⊥ = n i cos θ i − n t 1 − n i 2 n t 2 sin 2 θ i n i cos θ i + n t 1 − n i 2 n t 2 sin 2 θ i = n i cos θ i − n t 2 − n i 2 sin 2 θ i n i cos θ i + n t 2 − n i 2 sin 2 θ i = cos θ i − n t 2 n i 2 − sin 2 θ i cos θ i + n t 2 n i 2 − sin 2 θ i r ⊥ = n i cos θ i − n t 1 − n i 2 n t 2 sin 2 θ i n i cos θ i + n t 1 − n i 2 n t 2 sin 2 θ i = n i cos θ i − n t 2 − n i 2 sin 2 θ i n i cos θ i + n t 2 − n i 2 sin 2 θ i = cos θ i − n t 2 n i 2 − sin 2 θ i cos θ i + n t 2 n i 2 − sin 2 θ i

Similarly we can derive that

r ∥ = n t 2 n i 2 cos θ i − n t 2 n i 2 − sin 2 θ i n t 2 n i 2 cos θ i + n t 2 n i 2 − sin 2 θ i r ∥ = n t 2 n i 2 cos θ i − n t 2 n i 2 − sin 2 θ i n t 2 n i 2 cos θ i + n t 2 n i 2 − sin 2 θ i

t ⊥ = 2 cos θ i cos θ i + n t 2 n i 2 − sin 2 θ i t ⊥ = 2 cos θ i cos θ i + n t 2 n i 2 − sin 2 θ i

t ∥ = 2 n t n i cos θ i n t 2 n i 2 cos θ i + n t 2 n i 2 − sin 2 θ i t ∥ = 2 n t n i cos θ i n t 2 n i 2 cos θ i + n t 2 n i 2 − sin 2 θ i

This form allows us to easily plot the coefficients for different values of θ i . θ i . For example,here are the coefficients for the case where n t n i = 1.5 . n t n i = 1.5 .

Figure 1: The transmmission and reflection coefficients for the case where the ratio of transmitted to incident indices of refraction is 1.5. The top two curves are transmission. The lower two are reflection, with red being for the E field transverse to the plane of incidence.
Figure 1 (fresnel15.gif)
It is interesting to note that the sign of the coefficient can change on reflection. E r = − | r | E = e i π | r | E 0 e i ( K ⃗ ⋅ r ⃗ − ω t ) = | r | E 0 e i ( K ⃗ ⋅ r ⃗ − ω t + π ) E r = − | r | E = e i π | r | E 0 e i ( K ⃗ ⋅ r ⃗ − ω t ) = | r | E 0 e i ( K ⃗ ⋅ r ⃗ − ω t + π )

This corresponds to a phase change by π π upon reflection.

We can also look at what happens to the reflection coefficients when n i n t = 1.5 . n i n t = 1.5 .

Figure 2: The reflection coefficients for the case where the ratio of the incident to the transmitted incidence of reflection is 1.5.
Figure 2 (Fresnel2.gif)
That is going from a high index of refraction material to a lesser. In this case we see at the critical angle we get total internal reflection. What happens to the phase here is complicated.

When we have n i n t > 1 n i n t > 1 (or n t n i < 1 ) n t n i < 1 ) it is convenient to write r ⊥ = cos θ i − i sin 2 θ i − n t 2 n i 2 cos θ i + i sin 2 θ i − n t 2 n i 2 r ⊥ = cos θ i − i sin 2 θ i − n t 2 n i 2 cos θ i + i sin 2 θ i − n t 2 n i 2

Now to understand what this implies we need to digress a little. Recall that e i α = cos α + i sin α . e i α = cos α + i sin α . We could have written this as e i α = a + i b e i α = a + i b then we see that α = tan − 1 b a . α = tan − 1 b a . Now consider cos α − i sin α cos α + i sin α = e − i α e i α = e − 2 i α cos α − i sin α cos α + i sin α = e − i α e i α = e − 2 i α

Now looking back at r ⊥ r ⊥ we see that r ⊥ = e − i φ r ⊥ = e − i φ where tan φ 2 = sin 2 θ i − n t 2 n i 2 cos θ i . tan φ 2 = sin 2 θ i − n t 2 n i 2 cos θ i .

We could go through a similar excersize for r ∥ r ∥ and get the same result with

tan φ 2 = sin 2 θ i − n t 2 n i 2 n t 2 n i 2 cos θ i . tan φ 2 = sin 2 θ i − n t 2 n i 2 n t 2 n i 2 cos θ i . Figure 20-8 in Pedrotti and Pedrotti summarizes all the possible phase changes.

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