Today in Physics 218: Fresnel s equations

Transcription

1 Today in Physics 8: Fresnel s equations Transmission and reflection with E parallel to the incidence plane The Fresnel equations Total internal reflection Polarization on reflection nterference R Germanium n = θ ( degrees) E k, k, k E k, k, k R T R T 6 February 004 Physics 8, Spring 004

2 Last time: E perpendicular to the incidence plane for which we obtained E 0 E αβ = and E = E + αβ + αβ 0T 0R 0 for the transmitted and reflected amplitudes of E, where cosθt ε, and µ Z α = β = = cosθ ε µ Z Now, to complete the picture, we need to consider incident light with E polarized in the plane of incidence 6 February 004 Physics 8, Spring 004

3 E parallel to the incidence plane E R x µ, ε k R B R E T k T θ k θ θ T B T z E µ, ε B B and B point out of T the page; B points in R 6 February 004 Physics 8, Spring 004 3

4 E parallel to the incidence plane (continued) Again we will use the boundary conditions on E 0 cosθ + E 0R cosθ = E 0T cos θt, ( ) µε E µε E = µε E µ µ 0 0R 0T B This time it s the boundary condition that tell us nothing besides 0 = 0, and the D boundary condition that turns E 0 θ out to be identical to the sin θ E boundary condition E 0 cos θ θ E and H : k 6 February 004 Physics 8, Spring 004 4

5 E parallel to the incidence plane (continued) Now divide the first of these equations by cos θ, and divide the other one by ε µ : cos T E θ 0 + E 0R = E 0T = αe 0T, cosθ E µε 0 E 0R = E 0T = βe 0T, µε where, again, cosθ T µ and ε Z α = β cosθ = µ ε = Z t s a simple matter to solve these equations for the reflected and transmitted field amplitudes; in fact it s the same as we already did three times The result is: 6 February 004 Physics 8, Spring 004 5

6 E parallel to the incidence plane (continued) E 0 E α β 0T = and E 0R = E 0 α + β α + β All of these results comprise the Fresnel equations: E 0 αβ E k, kr, kt : E 0T =, E 0R = E 0 + αβ + αβ E 0 α β E k, kr, kt : E 0T =, E 0R = E 0 α + β α + β cosθt n α = sinθ cosθ cosθ n β = µ ε µ ε = Z Z ε ε = n n 6 February 004 Physics 8, Spring The applies for almost all nonconductors

7 Normal incidence We shall now turn to some interesting and useful implications of Fresnel s equations, and other relations that can be obtained in the same manner Normal incidence Both pairs of Fresnel equations reduce to the same expression we derived before for incidence angle zero ( α = cosθt cosθ = ): E 0 E β 0T = and E 0R = E 0, + β + β as they must if they re correct 6 February 004 Physics 8, Spring 004 7

8 Total reflection ( ) Suppose µ ε > µ ε n > n Then, at θ given by n sinθt = sinθc =, n the transmitted wave is parallel to the interface, and for values of θ> θc, we get sinθ T >! What does this mean? n this condition, n T = T = n cosθ sin θ sin θ n = i sin θ n < 0! 6 February 004 Physics 8, Spring 004 8

9 Total reflection (continued) Now the is real, but the projection of the wavevector is imaginary: i( kt r ωt) i( ktzcosθ+ ktxsinθ ωt E = E e = E e ) T 0T 0T = E n ktz sin θ n ikx ( T sinθ ωt) 0 T e e Real! The transmitted wave amplitude decreases exponentially with increasing z (exponential attenuation): no energy is transmitted to large distances, all energy is reflected, at incidence angles greater than θ C 6 February 004 Physics 8, Spring 004 9

10 Total reflection (continued) ( n = ) ( ) For example: flint glass 5 in air n = has n sinθc = = θc = 48 n 5 At θ > θc, light is totally reflected This is how light fibers work (Now you know how much a light fiber can be bent before it stops working) 6 February 004 Physics 8, Spring 004 0

11 Polarization on reflection f E 0 is parallel to the plane of incidence, then E α β 0R = E 0 = 0 at α = β ; α + β that is, all light polarized in the plane of incidence must be transmitted (none reflected) if this is true The incidence angle that corresponds to this is obtained from: cosθtb n α = = sinθb = β cosθb cosθb n n sinθb = β sin θ n ( ) B 6 February 004 Physics 8, Spring 004

12 Polarization on reflection (continued) n β B B n n β = β sin θ sin θ = β n Typically, µ = µ =, so β = n n : Compare these last two results: β β β β sin θb = = = ( β )( + β ) ( + β ) β β tan θ But, sinθ = tanθ cosθ = tan θ cos θ = + tan θ 6 February 004 Physics 8, Spring 004

13 Polarization on reflection (continued) n tan θb = β = n Note also that for θ = θb, n sinθ sin tan θ = θ = = n cosθ π sin θ sinθ = by Snell's law sinθt Thus θ + θt = π for light incident at Brewster s angle; the transmitted light travels perpendicular to the direction of reflection Brewster s angle µ ε µ, ε E, θ k E T θt k T 6 February 004 Physics 8, Spring 004 3

14 Polarization on reflection (continued) 08 Flint glass n = 5 08 Germanium n = 39 R R θ ( degrees) ( degrees) Fraction of incident intensity reflected, for materials in air E perpendicular to plane of incidence E parallel to plane of incidence 6 February 004 Physics 8, Spring θ

15 nterference You will show, in problem set #3, that the intensity transmitted by a plane parallel slab of dielectric material with refractive index n and thickness d, at normal incidence, is T = T = where k is the wavenumber in vacuum n + sinnkd n, Vacuum E k Vacuum E T kt n 6 February 004 Physics 8, Spring 004 5

16 nterference (continued) Look at this formula carefully and you ll see that T = (00% transmission) for certain values of k, T = for nkd = mπ, m= 0,,, because the sine term vanishes there The peaks get sharper with increasing refractive index n What is the origin of the peaks? (nterference, obviously, but how?) T nkd n = 5 n = 39 6 February 004 Physics 8, Spring 004 6

17 nterference (continued) One can think of the transmission as the result of a large number of internal reflections and transmissions At normal incidence, all of the transmitted waves would be in phase (and interfere constructively) if an integer number of wavelengths was covered for each two internal reflections: λ π d = m = m nkd = mπ, m= 0,,, n nk Just like the formula! 6 February 004 Physics 8, Spring 004 7

18 nterference (continued) This, too, can be solved as a boundary-value problem, by considering the boundary conditions at both surfaces, and supposing there are transmitted and reflected waves propagating inside the slab, as well as the incident and reflected waves on the incidence side, and just a transmitted wave on the far side The boundary conditions give us a system of four equations in four unknowns, in this case t can certainly be done this way, and we ll see an example ll also introduce a much handier way, that can be used easily to find transmission and reflection through an arbitrary number of dielectric slabs and surfaces 6 February 004 Physics 8, Spring 004 8