Obtain expressions for reflection and transmission amplitude coefficientswhenelectric vector associated with a plane monochromatic em waves in the plane of incidence.

Transverse waves oscillate in (y) perpendicular to the direction (positive x)
of propagation.  X-axis is along the string.  Let F be the tension
force in the string at any x.

     y(x, t) = Ai  Sin (ω t – k₁ x)    with an initial phase of 
0,  at  t = x = 0

             ω = angular
freq.  Wavelength = λ₁, wave
number = k₁ = 2π/λ₁ , 
            T =
time period 

            
velocity in x direction = v₁ = λ₁/T        k₁ = ω/v₁

let μ₁ = mass per
unit length of the string.   We can derive that :  v₁ = √(F/ μ₁)

So  F = μ₁ v₁² = v₁ Z₁, 
where Characteristic impedance of the string = Z₁ = μ₁ * v₁

      k₁ is the wave
number specific to the string and it may depend on the impedance of the string,
mass per unit length and temperature.   Suppose the wave encounters a
heavier string of higher wave number k₂, and impedance Z₂, then most of the energy in the wave is reflected. 
A little is transmitted.   The reflected wave has a phase difference
of π with the incident wave.  The transmitted wave has the same 
phase as the incident wave.   Frequency of the wave remains same in both
strings.

   Let  k₂, λ₂, v₂, Z₂  be the wave number, wavelength, velocity and
characteristic impedance of the wave on the second string.   Let both
strings meet at  x = L  and at   t = t₁.

                
So   v₂ = √(F/ μ₂)         and
    F = μ₂ v₂² = v₂  Z₂

                
Since, F = v₁ * Z₁ = v₂ *Z₂,  
we get   v₁ / v₂ = Z₂/Z₁ = k₂/k₁

 

      Yr (x, t) = Ar  Sin (ω (t-t₁) - k₁ (L- x) + θ)   where Ar = amplitude of
reflected wave. 

             As the
phase difference with Yi(x,t) is  π at x =L and t= t₁,  we get:

               
ω (t-t₁) - k₁(L-x) + θ = ω t₁ - k₁ L -
π       =>    θ = ω t₁ - k₁ L – π

         =>
      Yr(x,t) = Ar Sin [ω t - k₁(L- x) + k₁ L - π]

        Yt (x, t) = At  Sin [ω (t-t₁) - k₂ (x-L) + Ф)  , where At = amplitude of transmitted
wave

           as phase angle is
same as Yi(x,t) at  x = L and t = t₁ , we get

              ω (t-t₁) - k₂ (x-L) + Ф = ω t – k₁ x     => Ф = (k₂ – k₁) L 

           =>  Yt(x, t) = At Sin [[ω t -
k₂ x + (k₂  - k₁) L]

Boundary conditions :

1) Displacement of the initial wave is the algebraic sum of the other two
displacements at the boundary of the two strings.  It is like vector or
phasor addition.  Let δ be the phase angle of incident wave at the
boundary.

       Ai Sin δ  =  Ar Sin (δ - π) + At
Sin δ

                  
Ai   = At – Ar       =>  Ai +
Ar = At   --- (1)

2)   The energy incident at the boundary (per unit time) is split
into two components: reflected and transmitted.  Using the conservation of
energy principle, we get:

               
1/2  μ1 v1 ω^2  Ar^2 + 1/2 μ2 v2 ω^2 At^2 =  1/2  μ1 v1
ω^2  Ai^2 

                   => 
  Z1 (Ai^2 – Ar^2) = Z2  At^2

               
   =>   Z1 (Ai - Ar ) = Z2
At      --- (2)

Solving (1) and (2) we get:

               Ar = (Z1 – Z2) Ai  /
(Z₁ + Z₂)     
And     At = 2 Z₁ Ai /(Z₁ + Z₂ )

        Or,    Ar = (k₁ – k₂) Ai / (k₁ + k₂)     and     At
= 2 k₁ Ai /(k₁ + k₂)

       Or      Ar = (v₂ – v₁) Ai / (v₁ + v₂)     and      At
= 2 v₂ Ai /(v₁ + v₂)