obtain expression for reflection and transmission amplitude coefficients when electric vector associated with a plane monochromatic electromagnetic wave is in the plane of incidence.
Answers
A quick solution (assuming we know the symbols):
Ai = - Ar + At boundary
condition related to phase angles & displacement
Z₁ Ai² = Z₁ Ar² + Z₂ At² from Energy/sec. Conservation
or k₁ Ai² = k₁ Ar² + k₂ At²
Solving them we get
Ar/Ai = (k₁ - k₂)/(k₁+k₂) and At / Ai = 2 k₁ / (k₁+k₂)
or, Ar/Ai = (Z₁ - Z₂)/(Z₁+Z₂) and At / Ai = 2 Z₁/ (Z₁+Z₂)
or, Ar/Ai = (v₂ - v₁)/(v₁+v₂) and At / Ai = 2 v₂ / (v₁+v₂)
This is the same formula for all Transverse waves, mechanical (on a
string), light and electromagnetic waves too.
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Detailed derivation:
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)