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Reflection and Refraction. 11/07 05 n 1 n 2 For refraction at a plane surface between n 1 and n 2 E1r we have Eparallel and H (^) parallel continuous at the E 2 boundary. (All B's are out of the page) 1r 2 1 E 1 y z z = 0 At z = 0 we must match Eparallel and Hparallel at all y values. For example for E||, E 1 cos 1 exp( i k1y y) -E1r cos 1r exp( i kiry y) = E 2 cos 1 exp( i k2y y) This equation includes different angles for the reflected and transmitted waves, but in order to satisfy this equation (the time dependence has been omitted) at all y values on the boundary , absolutely must have k1y = k1ry = k2y Since k1y = k 1 sin 1 , and k1ry = k 1 sin 1r, settting them equal means 1 = 1r, the angle of incidence equals the angle of reflection. Now since k2y = k 2 sin 2 , setting k1y = k2y means k 1 sin 1 = k 2 sin 2. The angular frequency is the same on both sides, and k/ = 1/v, so we divide the last equation by : k 1 / sin 1 = k 2 / sin 2 , or sin 1 / v 1 = sin 2 / v 2 And from this since n = c/v we get snell's law: n 1 sin 1 = n 2 sin 2. Now we equate E|| at the boundary E 1 cos 1 - E1r cos 1 = E 2 cos 2 , or E 1 - E1r = (cos 2 )/( cos 1 ) E 2. For H|| we have H 1 + H1r = H 2 , or (E 1 + E1r)/( 1 v 1 ) = E 2 /( 2 v 2 )
With 1 = 2 = o, and n = c/v we have E 1 + E1r = (n 2 /n 1 ) E 2 Employing Griffiths notation of = (cos 2 )/( cos 1 ) and = v 1 /v 2 = n 2 /n 1 , we solve by adding the equations and get 2 E 1 = ( + ) E 2. The time averaged rate of energy flow is the poynting vector < S > = 1/2 Re ( ExH *). The transmission coefficient T is the ratio of outgoing power/area to the incoming power/area in the z direction T = z^ Re( E 2 xH 2 *)/ z^ Re( E 1 xH 1 *) Since for plane waves k 1 x E 1 = H 1 , E 1 x H 1 * = k 1 ( E E *)/() = k 1 ^ |E 1 |^2 /(v). Now the transmission coefficient can be written (taking the same on both sides) T = [Re{ (z^k 2 ^ ) |E 2 |^2 /v 2 } ]/[Re { (z^k 1 ^ ) |E 1 |^2 /v 1 ] , or T = Re(cos 2 )/ Re(cos 1 ) |E 2 /E 1 |^2 n 2 /n 1 = Re() |E 2 /E 1 |^2 When n 1 <n 2 , there are no problems for between 0 and 90o^ : is real and T = 4 /(+)^2 , or T = 4 (cos 2 / cos 1 ) (n 2 /n1) /{ cos 2 / cos 1 ) + n 2 /n 1 }. With cos 2 expressed in terms of 1 , and n 1 and n 2 , we are ready to plot, like Fig 9.17. At and beyond the critical angle. When n 1 >n 2 and 1 exceeds the 'critical angle', sin 2 >1, and is real. This will drive cos 2 to be pure imaginary as we will see in a moment, and T will be zero (no transmission beyond the 'critical angle'. We'll let 2 = a + ib, and write sin 2 = sin (a+ib) = sin a cos (ib) + cos a sin (ib). Since cos(ib) = cosh(b), and sin(ib) = i sinh(b), sin 2 = = sin a cosh (b) + i cos a sinh b. Sin 2 is real, so a must be /2 in order that the imaginary term vanish. Then sin 2 = cosh b. And then cos 2 is imaginary: cos 2 = (1- sin^2 2 ) = (1- cosh^2 b) = i sinh b. Then T = 0 because Re( cos 2 / cos 1 ) = 0 when n 1 >n 2 and 1 >c.
Subtracting them gives E1r = 1/2 [ ( 12 - 12 ) E 2 + ( 12 + 12 ) E2r ] We could regard these two as a matrix equation with column vectors 1 = (E 1 , E1r) and 2 = (E 2 , E2r) connected by a matrix [ ( 12 + 12 ) ( 12 - 12 ) ] 12 = 1/2 [ ] [ ( 12 - 12 ) ( 12 + 12 ) ] so that 1 = 12 2. When we go on to z = a to match the boundary conditions we get two more equations. These will involve E 2 and E2r, as well as E 3 and (if it existed) E3r. There would then be another matrix equation we could write down: 2 = 23 3. The two matrix equations could be joined, to give 1 = 12 2 = 12 ( 23 3 ) = ( 12 23 ) 3 = 13 3. The transmission coefficient in this situation involves E 3 and E 1. In particular T = Re(cos 3 / cos 1 ) n 3 /n 1 |E 3 /E 1 |^2. From the overall matrix equation we have E 1 = ( 13 ) 11 E 3 + ( 13 ) 12 E3r. Where there is no reflection, E3r = 0, and E 3 /E 1 = 1/( 13 ) 11 Thus it is possible to construct an overall matrix if for a multi-layered medium, and compute a transmision coefficient T = Re(cos f/ cos i) nf/ni /|(if) 11 |^2. This is an option for complicated situations, and we will not immediately use it here, just point out that there is power in the matrix approach. Now we will match things at z = a. For E 2 we will have E 2 = E2o exp ( i k2y y + i k2z a -it) = E2o exp ( i k2y y + i k 2 cos 2 a -it) There will be similar terms for E2r and for E 3 : E2r = E2ro exp ( i k2y y - i k2z a -it) E 3 = E2o exp ( i k3y y + i k3z a -it)
Notice that all y-terms match because of the boundary condition requirement (that gives us Snell's law) All terms have the same time dependence, so we only have to write down the z-part of the equation. Notice also that the z-component of k for the reflected wave E2r has a negative sign, because it moves to the left. Matching Eparallel at the boundary z =a we have cos 2 ( E 2 exp(i k2z a) - E2r exp (-ik2z a) ) = cos 3 E 3 exp( ik3z a) We could shorten this to read E 2 f 2 - E2r/f 2 = 23 E 3 f 3 using the notation 23 = cos 3 /cos 2 , f 2 = exp(i k2z a), and f 3 = exp(i k3z a). When we match Hparallel at z = a we find that (E 2 f 2 + E2r/f 2 )/(v 2 ) = E 3 f 3 /(v 3 ), which may be reduced to E 2 f 2 + E2r/f 2 = 23 E 3 f 3 Solving these for E 2 and E2r in terms of E 3 is done as before , by adding and subtracting. E 2 = 1/(2f 2 ) E 3 f 3 ( 23 + 23 ) , and E2r = f 2 /2 E 3 f 3 (- 23 + 23 ) Putting these together with E 1 = 1/2 [ ( 12 + 12 ) E 2 + ( 12 - 12 ) E2r ] , we find E 1 = E 3 f 3 /4 [ ( 12 + 12 ) ( 23 + 23 ) /f 2 + ( 12 - 12 ) f 2 (- 23 + 23 ) ]. where f 2 = exp(iF), F = k2z a, 12 = cos 2 /cos 1 , 23 = cos 3 /cos 2 , 12 =n 2 /n 1 , 23 =n 3 /n 2. Exercise: show that when F is real, T = n 3 /n 1 |E 3 /E 1 |^2 = 4 n 3 /n 1 /{ ( 12 23 + 23 12 )^2 + ( 232 - 232 )( 122 - 122 ) cos^2 (F) } Exercise: For normal incidence, show that T=1 when F = /2 and n 22 = n 1 n 3. (Non-reflecting coating.)
