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