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Transcript
S. Blair September 27, 2010 9 2.4. Wave propagation at an interface - reflection and refraction Now, if we take the plane wave solution to the wave equation, we can determine what happens when a wave strikes a dielectric interface. We know from our discussion on ray optics that when a ray hits a mirror, the angle of reflection (with respect to the normal) equals the angle of incidence. Also, we know that for a transmitted ray through a dielectric interface, the angle of refraction and angle of incidence are related by Snell’s Law ni sin θi = nt sin θt , where ni is the refractive index of the incidence medium, nt is the refractive index of the transmission medium, and θ are the respective angle with respect to the normal at the interface. What ray optics doesn’t tell us is that at a dielectric interface, we will in general have both transmitted and reflected components. In order to determine the amplitudes of the refracted and reflected components, we have to use the wave optics description with plane waves. The incident, reflected, and transmitted plane wave components can be written Ei = Eio ej(ωt−ki ·r) Er = Ero ej(ωt−kr ·r) Et = Eto ej(ωt−kt ·r) The field amplitudes Eio , etc., can represent either the TE (polarization perpendicular to plane of incidence E⊥ ) or TM (electric field polarization in the plane of incidence E|| ) polarizations. Similar equations can be written for the magnetic fields, where B⊥ = (n/c)E|| and B|| = (n/c)E⊥ . E t,// y θt z x into paper ki E i,// θi θ r E i,⊥ Incident wave E r,// Transmitted wave kt E t,⊥ E r,⊥ kr Reflected wave The next thing we have to specify are the boundary conditions, which state that the tangential components of the electric and magnetic fields are continuous across an interface. This leads to the following relationships for the TE wave: Ei,⊥ + Er,⊥ = Et,⊥ Bi,|| cos θi + Br,|| cos θr = Bt,|| cos θt , which states that the x component of the electric field and z component of the corresponding magnetic field are continuous. This leads to the following equations: Eio,⊥ ej(ωt−kiz z) + Ero,⊥ ej(ωt−krz z) = Eto,⊥ ej(ωt−ktz z) Bio,|| cos θi ej(ωt−kiz z) − Bro,|| cos θr ej(ωt−krz z) = Bto,|| cos θt ej(ωt−ktz z) . S. Blair September 27, 2010 10 Since these equations must be valid for all values of t and z, we must have that (ωt − kiz z) = (ωt − krz z) = (ωt − ktz z) , Since the frequency ω is the same across the interface, the z wavevector components must satisfy kiz = krz = ktz ⇒ n1 sin θi = n1 sin θr = n2 sin θt , which give the laws of reflection and refraction. Canceling the exponential terms, the amplitudes are related by Eio,⊥ + Ero,⊥ = Eto,⊥ Bio,|| cos θi − Bro,|| cos θr = Bto,|| cos θt . Now, using the relationship between the electric and magnetic field amplitudes, Eio,⊥ + Ero,⊥ = Eto,⊥ n1 Eio,⊥ cos θi − n1 Ero,⊥ cos θr = n2 Eto,⊥ cos θt , we can derive the reflection and transmission coefficients for the TE polarization Er0,⊥ cos θi − n2 − sin2 θi r⊥ = = Ei0,⊥ cos θi + n2 − sin2 θi Et0,⊥ 2 cos θi t⊥ = = Ei0,⊥ cos θi + n2 − sin2 θi where n = n2 /n1 . We can show that the following additional relationship holds: r⊥ + 1 = t ⊥ . Following the same steps for the TM polarization, we get Er0,|| n2 − sin2 θi − n2 cos θi r|| = = Ei0,|| n2 − sin2 θi + n2 cos θi Et0,|| 2n cos θi t|| = = Ei0,|| n2 − sin2 θi + n2 cos θi 1 = r|| + nt|| . S. Blair September 27, 2010 11 2.4.1. Reflection amplitude and phase - n1 > n2 The solid line is TM polarization, and dotted line is TE polarization. Brewster’s angle occurs when n2 tan θp = . n1 At this angle, r|| = 0, so that the TM polarization is completely transmitted. Brewester’s angle is used in may different lasers to force the output to be in one state of polarization. Total internal reflection occurs when sin θc = n2 , n1 and only happens when n1 > n2 . Under TIR, both the TE and TM waves are completely reflected back into the incidence medium. This effect is used extensively in optical waveguides. 2.4.2. Reflection amplitude and phase - n1 < n2 S. Blair September 27, 2010 12 2.4.3. Reflectance and transmittance The quantity reflectance is the ratio of the reflected intensity to the incident intensity: R⊥ = likewise |Er0,⊥ |2 = |r⊥ |2 , 2 |Ei0,⊥ | R|| = |r|| |2 . At normal incidence (i.e. θi = 0), these expression reduce to R⊥ = R|| = n1 − n2 n1 + n2 2 . For air to glass with n1 = 1 and n2 ∼ 1.5, R ∼ 4%. The transmittance is the ratio of transmitted intensity to incident intensity: T⊥ = n2 |t⊥ |2 n1 T|| = n2 2 |t|| | , n1 for normal incidence. The reason for the refractive index ratio is that intensity is defined as I = |E|2 /2η = v0 r |E|2 /2, and r (and therefore n and η) are different for the incident and transmitted waves. More generally, in the absence of absorption, the transmittance can be calculated from T = 1 − R.