// Main VL template #import "../preamble.typ": * // Fix theorems to be shown the right way in this document #import "@preview/ctheorems:1.1.3": * #show: thmrules // Main settings call #show: conf.with( // May add more flags here in the future num: 5, type: 0, // 0 normal, 1 exercise date: datetime.today().display(), //date: datetime( // year: 2025, // month: 5, // day: 1, //).display(), ) = Uebersicht == Reflexion und Brechung Wechselstroeme und das Zeigerdiagramm $ I (t) = I_0 cos [omega t + phi_(I) ] \ U (t) = U_0 cos [ omega t + phi _(U) ] \ Z = a + i b = abs(z) exp(i phi). $ Wiederholung von Kombinantion von Wechselstromwiederstaenden und deren Impedanz $ abs(Z) = sqrt(R^2 + (omega L - 1/ (omega C))^2 ) \ tan phi = (omega L - 1/(omega C)) / (R) = (Im Z) / (Re Z). $ Es gilt $ (diff V) / (diff x) = - ((diff L) / (diff x) )(diff I) / (diff t) \, space (diff I) / (diff x) = - ((diff C) / (diff x) )(diff V) / (diff t) \ (diff ^2 V) / (diff t^2 ) = - 1/C_0 diff / (diff x) (diff I) / (diff t) = - 1/C_0 diff / (diff x) (- 1/L_0 (diff V) / (diff x) )= 1 / (L_0 C_0 )(diff ^2 V) / (diff x^2 ) \, space c = 1/sqrt(L_0 C_0 ) \ partial _(x) I = - C_0 partial _(t) V ==> I_(+) = I_(0 +) e^(i (k x - omega t)) \, space V_(+) = V_(0 +) e^(i (k x - omega t)) \ I_(0 +) i k = - C_0 (i omega) V_(0 +) \ ==> (V_(0 +) ) / (I _(0 +) ) = k/omega 1/C_0 = 1/c 1/C_0 = sqrt(L_0 C_0 )1/C_0 = sqrt((L_0 ) / (C_0 ) ) = Z_0. $ = Koaxialkabel Es gilt $ E = 1/(2 pi epsilon_0 ) (lambda) / (r) = 1/(2 pi epsilon_0 ) (Q) / (L r) \ Delta V = V_(b) - V_(a) \ dif C = (2 pi epsilon_0 ) / (ln (R/r)) dif l \ dif L = (mu mu_0 ) / (2 pi) ln (R/r) dif l. $ Wir haben Wellen der Form $ I, U, arrow(E), arrow(B). $ Fuer die Impedanz gilt dann $ Z := sqrt(L/C) \, space [Z]= Omega. $ Die Grenzflaeche ist dann zwischen $Z_(1) $ und $Z_2 $. Es gilt fuer die drei Stroeme $ I = I_(1 + ) + I _(1 - ) = I _(2 + ) \ U = U_(1 + ) + U_(1 - ) = Z_(1) I _(1 + ) - Z_(1) I_(1 - ) = U _(2 + ) = Z_(2) I_(2 + ) \ ==> I_(2 + ) = I _(1 + ) + I _(1 - ) \, space Z_(2) I_(2 + ) = Z_(1) I _(1 + ) + Z_(1) I _(1 - ) \ ==> 1 + (I_(1 - ) ) / (I _(1 + ) ) = (I _(2 + ) ) / (I _(1 + ) ) := underbrace(t, "Transmission") = 1 + underbrace(r, "Reflexion") \ Z_(1) - Z_(2) (I_(1 - ) ) / (I _(1 + ) ) = Z_(2) (I_(2 + ) ) / (I _(1 + ) ) \ ==> r = (Z_(1) - Z_(2) ) / (Z_(1) + Z_(2) ) \, space t = (2 Z_(1) ) / (Z_(1) + Z_(2) ). $ = Dopplereffekt Fuer Schall $ f' = (1 +- v_(e) /c) / (1 minus.plus v_(s) /c) f_0, $ wobei $+ $ aufeinander zu und $- $ voneinander weg ist. Auch ist $v_(e) $ die Geschwindigkeit des Empfaengers und $v_(s) $ die des Senders. Lorentztransformation impliziert $ f = sqrt((c +- v) / (c minus.plus v) ) f_0. $ Fuer die Wellenlaenge $lambda$ $ lambda = c T \, space f_0 = c /T \ lambda' = (c + v_(s) ) T \, space lambda'' = (c - v_(s) ) T \ f' = c/lambda' = (c) / ((c + v_(s) )T) = c / (v + v_(s) )f_0 ==> f' = 1/(1 +- v_(s) /c) f_0. $ Fuer den bewegten Empfaenger $ T' = (lambda) / (c + v_(e) ) = T_0 /(1 + v_(e) /c) \ ==> f'= (1 + v_(e) /c) f_0 \ f' = (1 +- v_(e) /c) / (1 minus.plus v_(s) /c) f_0. $ Fuer Relativitaet $ x' = gamma (x - v t) \, space y' = y \, space z' = z \, space t' = gamma (t - (x v) / (c^2 ) ) \, space gamma = (1) / (sqrt(1 - v^2 /c^2 )) \ ( c Delta t - v Delta t)/N \, space f = c/lambda = (c N) / (c Delta t - v Delta t) \ N = f_0 Delta t' \, space Delta t' = Delta t = (Delta t)/gamma \ f = (c) / (c Delta t - v Delta t) f_0 (Delta t) / (gamma) = 1/ (1 - v/c) f_0 /gamma = sqrt((1 + v/c)/(1 - v/c)) f_0 $ = Brechung und Reflexion Betrachte Wellen $ psi_(1) = a_1 e ^(i (arrow(k)_(1) arrow(r) - omega t)) \ psi_(r) = a_(r) e^(i (arrow(k)_(r) arrow(r) - omega _(r) t)) psi_(2) = a_(2) e^(i (arrow(k)_(2) arrow(r) - omega _(2) t)). $ Das Ziel ist die Bestimmung der Geometrie. Also $Theta_(2) , Theta_(1) "und" Theta_(r) $ und $R = abs((a_(r) ) / (a_(1) ) )^2 \, space R = R (n_1, n_2, Theta_(1) )$ sowie $T = abs(a_2 /a_1 )^2 \, space T = T (n_1, n_2, Theta_(1) ) $. Es gilt dabei $R = abs(r)^2 \, space T = abs(t)^2 $. Alles folgt aus den Randbedingungen. So ist $psi$ hier stetig und $ arrow(nabla) psi "stetig an der Grenzflaeche" z = 0. $ Stetigkeit von $psi$ impliziert $ a_1 e^(i (arrow(k)_(1) arrow(r) - omega t)) + a_(r) e^(i (arrow(k)_(r) arrow(r) - omega_(r) t)) = a_(2) e^(i (arrow(k)_(2) arrow(r) - omega_(2) t)) space forall arrow(r) = vec(x, y, 0) space forall t \ ==> omega := omega_(1) = omega_(r) = omega_(2). $ Es muessen also auch die Tangentialkomponenten von $arrow(k)$ gleich sein, also gilt auch $ k_(1) ^((x)) = k_(r) ^((x)) = k_(2) ^((x)). $ Also $ k_(1) ^((x)) = k n_(1) sin Theta_(1) = k n_(1) sin Theta_(r) \ ==> Theta_(1) = Theta_(r) "Reflexionsgesetz". $ Aus $k_(1) ^((x)) = k_(2) ^((x)) $ folgt $ k n_(1) sin Theta_(1) = k_(2) n_(2) sin Theta_(2) \ ==> n_1 / n_2 = (sin Theta_(2) ) / (sin Theta_(1) ). $ Mit $k = (2 pi )/lambda$ im Vakuum. Auch $ a_1 + a_(r) = a_(2). $ Aus der Stetigkeit des Gradienten folgt $ arrow(k)_(1) psi_(1) + arrow(k)_(r) psi_(r) = arrow(k)_(2) psi_(2) \ ==> k n_1 a_1 sin Theta_(1) + k n_1 a_1 sin Theta_(1) = k n_2 a_2 sin Theta_2 \ ==> (a_1 + a_(2) ) n_1 sin Theta_1 = a_2 n_2 sin Theta_2 "wieder der Snellius". $ Die senkrechte Komponenente liefert $ - k n_1 a_1 cos Theta_1 + k n_1 a_(r) cos Theta_1 = - k n_2 a_2 cos Theta_2 \ ==> n_1 cos Theta_1 (a_1 - a_(r) ) = n_2 (a_1 + a_(r) )cos Theta_2 \ ( 1 - a_(r) /a_(1) ) = n_2 /n_1 (cos Theta_2 )/(cos Theta_1 ) (1 + a_(r) /a_(1) ) "mit" r := a_(r) /a_(1) ==> (1 - sigma) = n_2 /n_1 (cos Theta_2 ) / (cos Theta_1 ) space "?" $ Also $ 1 - n_2 /n_1 (cos Theta_2 ) / (cos Theta_1) = r (n_2 /n_1 (cos Theta_(2) ) / (Theta_(1) ) + 1) \ ==> r = (n_2 cos Theta_1 - n_2 cos Theta_2 ) / (n_1 cos Theta_(1) + n_2 cos Theta_(2) ) \ ==> t = a_(2) /a_(1) = (a_1 + a_(r) ) / (a_1 ) = 1 + r \ ==> t = (2 n_1 cos Theta_1 ) / (n_1 cos Theta_1 + n_2 cos Theta_2 ). $ Jetzt als Spezialfall den senkrechten Einfall $ Theta_1 = 0 degree \, space r = (n_1 - n_2 ) / (n_1 + n_2 ) \ "fuer" n_1 = 1 \, space n_2 = 1.5 space ("Glas") \ ==> r = (- 0.5) / (2.5) = - 0.2 \, space R := r^2 = 0.04. $ Eine Grenzflaeche mit Normalen in zwei Medien mit Brechungsindizes $n_1 "und" n_2 $ $ Theta_1 = Theta_(c) <==> Theta_(2) = 90 degree. $ Ist der Snellius fuer $Theta > Theta_(c) $ erfuellbar? Gibt es Verhaeltnisse fuer die der Reflexionskoeffizient $R = abs(r)^2 = 1$ sein kann? #highlight[TODO: Totalreflexion herleiten]