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

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one year ago

```							Transverse waves oscillate in (y) perpendicular to the direction (positive x)of propagation.  X-axis is along the string.  Let F be the tensionforce 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 = angularfreq.  Wavelength = λ₁, wavenumber = k₁ = 2π/λ₁ ,             T =time period             velocity in x direction = v₁ = λ₁/T        k₁ = ω/v₁let μ₁ = mass perunit 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 wavenumber specific to the string and it may depend on the impedance of the string,mass per unit length and temperature.   Suppose the wave encounters aheavier 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 differenceof π with the incident wave.  The transmitted wave has the same phase as the incident wave.   Frequency of the wave remains same in bothstrings.   Let  k₂, λ₂, v₂, Z₂  be the wave number, wavelength, velocity andcharacteristic impedance of the wave on the second string.   Let bothstrings 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 ofreflected wave.              As thephase 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 transmittedwave           as phase angle issame 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 twodisplacements at the boundary of the two strings.  It is like vector orphasor addition.  Let δ be the phase angle of incident wave at theboundary.       Ai Sin δ  =  Ar Sin (δ - π) + AtSin δ                  Ai   = At – Ar       =>  Ai +Ar = At   --- (1)2)   The energy incident at the boundary (per unit time) is splitinto two components: reflected and transmitted.  Using the conservation ofenergy principle, we get:               1/2  μ1 v1 ω^2  Ar^2 + 1/2 μ2 v2 ω^2 At^2 =  1/2  μ1 v1ω^2  Ai^2                    =>   Z1 (Ai^2 – Ar^2) = Z2  At^2                  =>   Z1 (Ai - Ar ) = Z2At      --- (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)
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one year ago
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