// 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]
