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Radiation perpendicularly incident to a surface

  1. Nov 17, 2011 #1
    1. The problem statement, all variables and given/known data
    Consider radiation perpendicularly incident to a plane whose index of refraction is n. Show that the reflected wave can be eliminated by covering the plane with a second dielectric layer with index of [itex]\sqrt{n}[/itex] and thickness 1/4 wavelength.


    2. Relevant equations
    Boundary conditions: [itex]E_{1}^{||}=E_{2}^{||}[/itex], and [itex]B_{1}^{||}=B_{2}^{||}[/itex]


    3. The attempt at a solution

    Electric and Magnetic field at z=0 can be given by:

    [itex]{E_{I}(z,t)=E_{I}exp[i(k_{1}z-\omega t)],E_{R}(z,t)=E_{R}exp[i(-k_{1}z-\omega t)]},{B_{I}(z,t)=\frac{1}{v_{1}}E_{I}exp[i(k_{1}z-\omega t)],B_{R}(z,t)=\frac{1}{v_{1}}E_{R}exp[i(-k_{1}z-\omega t)]}[/itex]

    For 0<z<[itex]\frac{\lambda}{4}[/itex]:

    [itex]{E_{i}(z,t)=E_{i}exp[i(k_{2}z-\omega t)],E_{r}(z,t)=E_{r}exp[i(-k_{2}z-\omega t)]},{B_{i}(z,t)=\frac{1}{v_{1}}E_{i}exp[i(k_{2}z-\omega t)],B_{r}(z,t)=\frac{1}{v_{1}}E_{r}exp[i(-k_{2}z-\omega t)]}[/itex]

    For z>[itex]\frac{\lambda}{4}[/itex]:

    [itex]{E_{T}(z,t)=E_{T}exp[i(k_{3}z-\omega t)]},{B_{T}(z,t)=\frac{1}{v_{1}}E_{I}exp[i(k_{3}z-\omega t)]}[/itex]

    These boundary conditions allow us to find simultaneous equations for z=0 and z=[itex]\frac{\lambda}{4}[/itex]:

    z=0 => [itex]{E_{I}+E_{R}=E_{r}+E_{i}, E_{I}-E_{R}=\beta (E_{r}-E_{i})}[/itex]

    z=[itex]\frac{\lambda}{4}[/itex]=> [itex]{E_{r}exp[ik_{2}\frac{\lambda}{4}]+E_{i}exp[-ik_{2}\frac{\lambda}{4}]=E_{T}exp[ik_{3}\frac{\lambda}{4}], E_{r}exp[ik_{2}\frac{\lambda}{4}]-E_{i}exp[-ik_{2}\frac{\lambda}{4}]=\alpha E_{T}exp[ik_{3}\frac{\lambda}{4}]}[/itex]

    Here, [itex]\alpha =\frac{v_{2}}{v_{3}}[/itex], [itex]\beta =\frac{v_{1}}{v_{2}}[/itex]

    I have done plenty of algebra to find [itex]E_{I}[/itex] in terms of [itex]E_{R}[/itex], but this seems to require more information than I have. I am just not seeing the end from here, does anyone have a suggestion on where to go from this point?
     
  2. jcsd
  3. Nov 18, 2011 #2
    I can offer only limited help !!! I do not know where √n comes into the full mathematical analysis BUT, at my level, it is easy to show that the film needs to be λ/4 thick.
    Here goes: Light from the low n to high n interface undergoes a phase shift of ∏ (λ/2)
    Light striking the high n to low n interface does not undergo a phase shift.
    In this example BOTH reflections undergo a λ/2 phase shift because both are from low n to high n.
    To eliminate the reflected wave needs interference between the 2 reflected waves and the phase shift must be λ/2 (or 3λ/2, 5λ/2 ....etc).
    This means that the film must be λ/4 thick ...light travels there and back through the film.
    That is the level of my understanding !!!!
    Hope it helps
     
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