Today in Physics 218: impedance of the vacuum, and Snell s Law
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1 Today in Physics 218: impedance of the vacuum, and Snell s Law The impedance of linear media Spacecloth Reflection and transmission of electromagnetic plane waves at interfaces: Snell s Law and the first Fresnel relation Lots of small pieces of Eccosorb spacecloth (Emerson & Cuming Microwave Products Corp.) 4 February 2004 Physics 218, Spring
2 Electromagnetic impedance of linear media Enclose a hypothetical source of electromagnetic radiation with a surface. The power flowing through the surface will be c c P() t = ( ) ( ˆ S da= E H da= E µε k E) da 4π 4πµ c ε c ε = = 4π µ 4π µ ( ) ( ˆ ( ˆ )) 2 ke E E k E da 2 2 Veff 1d 2 c ε = E d = 4π µ 4π µ µ µ r µ 0 wher e Z = = = in MKS units c ε ε εrε0 4 February 2004 Physics 218, Spring Z, E da
3 Electromagnetic impedance of linear media (continued) In vacuum, 4π 10-1 Z = Z0 = = sec cm in cgs units, c µ 0 = = Ω in MKS units. ε 0 (Officially, ohms per square, as we ll see shortly.) As in the case of the string, the amplitude ratios for light reflected and transmitted by the interface between two linear media can be expressed compactly in terms of impedance, because ε 2µ 1 Z β = = 1 : ε µ Z February 2004 Physics 218, Spring
4 Electromagnetic impedance of linear media (continued) 2 2Z E E E 2 0T = 0I = 0I 1 + β Z2 + Z1 1 Z Z E β E E 2 1 0R = 0I = 0I 1 + β Z2 + Z1 Note that the amplitude ratios come out simple in terms of impedance, even if µ 1. Note also that there s no reflected light if the impedances of the two media are equal (impedances matched). This will be true if ε 2µ 1 = ε 1µ 2; the media can be quite different and still have the same impedance.,. 4 February 2004 Physics 218, Spring
5 Spacecloth What does it mean for vacuum to have an impedance of 377 Ω? (Most resistors have terminals!) Well, consider an infinite slab of material with resistivity ρ and thickness d, and consider further a square of side L within the slab. For voltage difference V between two sides of the square, a current will encounter a L resistance ρl ρl ρ R = = =. A Ld d L This is true for any square, any size, any orientation. d 4 February 2004 Physics 218, Spring
6 Spacecloth (continued) Space is like this: any square (any size) taken from a slab of vacuum exhibits an impedance for electric fields applied across the square. One may think of this impedance as originating in the induction of E and B from each other: try to change E, for instance, and energy has to be put into B as well. A medium with resistivity and thickness such that its resistance per square is the same as vacuum would absorb light without reflection; that is, it would look to incident light as if it were an infinite vacuum. Such material exists: microwave engineers call it spacecloth. 4 February 2004 Physics 218, Spring
7 Spacecloth (continued) One prominent example of the use of highly optimized, durable spacecloth. (USAF photo.) 4 February 2004 Physics 218, Spring
8 Reflection and transmission for oblique incidence Now let s consider plane electromagnetic waves incident obliquely on an infinite, planar surface between two different linear media, a case which has no counterpart in the realm of strings. The geometry is shown on the next page, and the fields are i( ki r ωt E ) ˆ I = E 0Ie B I = µε 1 1kI E I i( kr r ωt E ) ˆ R = E 0Re B R = µε 1 1kR E R i( kt r ωt E = E e ) B = µε kˆ E T 0T T 2 2 T T and our problem is to find E R, E T, kr, and kt in terms of E I and ki. The former two will work out from the boundary conditions; the latter just from the fact that we are applying boundary conditions. 4 February 2004 Physics 218, Spring
9 Reflection and transmission for oblique incidence (continued) kr x kt θ R θ I θ T z k I 1, 1 µ ε µ 2, ε2 The ks don t necessarily lie in a plane: we have to prove this. 4 February 2004 Physics 218, Spring
10 Reflection and transmission for oblique incidence (continued) Solution: At z = 0, write the continuity conditions: all four this time: ε1 E,1 ε2 E,2 = 0 B,1 B,2 = E,1 E,2 = 0 B,1 B,2 = 0 µ or, 1 µ 2 ε E + E = ε E B + B = B ( ) 1 0Iz 0Rz 2 0Tz 0Iz 0Rz 0Tz 1 1 E + E = E B + B = B ( ) 0Ix 0Rx 0Tx 0Ix 0Rx 0Tx µ 1 µ 2 1 ( ) 1 E + E = E B + B = B µ 1 µ 2 All of these are of the form i( ki r t) i( R t) i( T t Ae ω k r Be ω k r + = Ce ω ). 0Iy 0Ry 0Ty 0Iy 0Ry 0Ty 4 February 2004 Physics 218, Spring
11 Reflection and transmission for oblique incidence (continued) As we ve seen, this sort of result is not special to electromagnetic waves, but appears in boundary conditions for any sort of wave. Now, it is an interesting fact that, for nonzero constants A, B, C, a, b, and c, if iau ibu icu Ae + Be = Ce then a = b = c and A + B = C, for all u, as you will show in this week s homework (problem 9.15). For our boundary conditions, this means ki r = kr r = kt r at z = 0, or k x+ k y = k x+ k y = k x+ k y. Ix Iy Rx Ry Tx Ty, 4 February 2004 Physics 218, Spring
12 Reflection and transmission for oblique incidence (continued) Furthermore, along the x axis, kixx = krxx = ktxx, and similarly kiyy = kryy = kty y along the y axis. If we were to orient the axes such that k Iy = 0, then kry = kty = 0 as well, and the situation will look exactly like the picture a couple of slides ago, with all the kslying in a plane. Thus there is always such a plane: ki, kr, and kt lie in a plane. This plane is called the plane of incidence. 4 February 2004 Physics 218, Spring
13 Reflection and transmission for oblique incidence (continued) Now back to the figure. From kixx = krxx = ktxxat z = 0 we also get ki sinθi = krsinθr = kt sin θt. But ki = kr = ω v1 = ωn1 c, so the first equation gives us the familiar mirror reflection law: sinθi = sin θr θi = θr. And k T = ω v 2, so sinθt k v2 n1 sinθ I k v n. Snell s Law I T 1 2 Since the conditions arise just from continuity, rather than the physics that gives rise in this case to continuity, these laws will apply to waves quite generally. 4 February 2004 Physics 218, Spring
14 Reflection and transmission for oblique incidence (continued) Since the wave vectors all lie in the incidence plane and the fields are transverse, we can treat separately the cases of light with E polarized perpendicular to, or parallel to, the plane of incidence. First we ll treat the perpendicular case, also known as problem! February 2004 Physics 218, Spring
15 E perpendicular to incidence plane B R x µ 2, ε2 k R E E I R k I θ I θ I θ T E T B T k T z µ 1, ε1 B I All Es point up, out of the page. 4 February 2004 Physics 218, Spring
16 E perpendicular to incidence plane (continued) Let s use the same two boundary conditions that we used k I θ I for normal incidence, continuity of the parallel components. θ I (We only need two, because B 0I cosθi E 0R and E 0T are the only unknowns left.) At z = 0, B 0I sin θi these are E 0I + E 0R = E 0T, 1 ( ) 1 µ 1ε1E 0I cosθ I + µ 1ε1E 0R cosθ I = µ 2ε2E 0T cos θt. µ µ 1 2 The D condition tells us nothing this time; the B one tells us nothing new: 4 February 2004 Physics 218, Spring
17 E perpendicular to incidence plane (continued) Using Snell s Law on this, n we get µε E sinθ + µε E sinθ = µε E sin θ I I 1 1 0R I 2 2 0T T sinθ = n sinθ µ ε sinθ = µ ε sin θ, 1 I 2 T 1 1 I 2 2 T E E E 0I + 0R = 0 T, which we already knew. We will generally find this to be true in our upcoming work: one boundary condition will tell us nothing, and two of the remaining ones will be equivalent. Also generally, though, the parallel-component BCs tell us different things, so well use them a lot. 4 February 2004 Physics 218, Spring
18 E perpendicular to incidence plane (continued) cosθ Now define T ε 2µ 1 Z α =, β = = 1, cosθi ε 1µ 2 Z2 and divide the H equation by µ 1ε1 cos θ I : E 0I + E 0R = E 0T, E E = αβe, 0I 0R 0T almost like the string again. Add them to get 2E0I 2E 0I = ( 1 + αβ ) E 0T E 0T =. 1 + αβ Multiply the first one by αβ and subtract to get 1 αβ 1 E 0I E 0R = 0 E 0R = E 0I. 1 + αβ ( αβ ) ( αβ ) Same as before, if α = 1. 4 February 2004 Physics 218, Spring
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