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[thin films] (Macleod's) phase-shift upon reflection

  1. Apr 14, 2008 #1
    Folks,

    I'm reading the 3rd edition of Thin Film Optical Filters by Macleod, and I've derived his expression for phase shift upon reflection (page 45). Finally, it lead me to a question. Any help is highly appreciated!

    [tex]\rm E_i + E_r = E_t \qquad (1)[/tex]
    [tex]\rm H_i - H_r = H_t \qquad (2)[/tex]

    Rewrite (2) using the definition of tilted optical admittance:

    [tex]\rm \eta_0 E_i - \eta_0 E_r = \eta_1 E_t \qquad (3)[/tex]

    Substitute (1) into (3):

    [tex]\rm \eta_0 E_i - \eta_0 E_r = \eta_1 \left( E_i + E_r \right) \qquad (4)[/tex]

    Rearrange (4):

    [tex]\rm E_i \left( \eta_0 - \eta_1 \right) = E_r \left( \eta_0 + \eta_1 \right) \qquad (5)[/tex]

    Apply the definition of reflection coefficient [tex]\rho[/tex]:

    [tex] \rho \equiv \rm \frac{E_r}{E_i} = \frac{ \eta_0 - \eta_1}{ \eta_0 + \eta_1} \qquad (6)[/tex]

    Replace [tex]\eta_1[/tex] by the input optical admittance [tex]Y[/tex] in (6):

    [tex] \rho = \frac{ \eta_0 - Y}{ \eta_0 + Y} \qquad (7)[/tex]

    By definition, [tex]Y \equiv C/B[/tex]. Then

    [tex]\rho = \frac{ \eta_0 - C/B}{ \eta_0 + C/B} = \frac{ \eta_0 B - C}{ \eta_0 B + C} \qquad (8)[/tex]

    From the general form of the characteristic matrix

    [tex]\begin{bmatrix}
    \displaystyle B \\
    \displaystyle C
    \end{bmatrix} =
    \begin{bmatrix}
    \displaystyle \alpha & \displaystyle i\beta \\
    \displaystyle i\gamma & \displaystyle \delta
    \end{bmatrix}
    \begin{bmatrix}
    \displaystyle 1 \\
    \displaystyle \eta_m
    \end{bmatrix} =
    \begin{bmatrix}
    \displaystyle \alpha + i\beta \eta _m \\
    \displaystyle \delta \eta_m + i \gamma
    \end{bmatrix} \qquad (9)
    [/tex]

    let [tex]B = \alpha + i\beta \eta _m[/tex] and [tex]C = \delta \eta_m + i \gamma[/tex] in (8).

    (problems previewing all of this latex code in one post - I'll break it up - please wait)
     
  2. jcsd
  3. Apr 14, 2008 #2
    OK. Part II...

    Therefore

    [tex]\rho = \frac{\eta_0 \left( \eta_m ^2 \beta^2 + \alpha ^2 \right) - \eta _m ^2 \delta ^2 - \gamma ^2 + i \, 2 \eta_0 \left( \eta_m ^2 \beta \delta - \alpha \gamma \right)}{\eta_0 \left( \eta_m ^2 \beta^2 + \alpha ^2 \right) + 2 \eta_0 \eta_m \left( \alpha \delta + \beta \gamma \right) + \eta _m ^2 \delta ^2 + \gamma ^2} \qquad (10) [/tex]

    so

    [tex]\varphi = \arctan \left| \frac{2 \eta_0 \left( \eta_m ^2 \beta \delta - \alpha \gamma \right)}{\eta_0 \left( \eta_m ^2 \beta^2 + \alpha ^2 \right) - \left( \eta _m ^2 \delta ^2 + \gamma ^2 \right)} \right| \qquad (11)[/tex]

    By inspection of (11), it follows

    [tex]\varphi = \arctan \left| \frac{\eta_0 \Im \left\{ B C ^{\ast} - CB^{\ast} \right\} }{\eta_0 ^2 BB^{\ast} - CC^{\ast}} \right| \qquad (12)[/tex]

    Finally, if [tex]\eta_0 = \eta_m[/tex], I arrive at Macleod's expression for phase-shift upon reflection

    [tex]\varphi = \arctan \left| \frac{\eta_m \Im \left\{ B C ^{\ast} - CB^{\ast} \right\} }{\eta_m ^2 BB^{\ast} - CC^{\ast}} \right| \qquad (13)[/tex]

    Note: [tex]\eta_m[/tex] is the tilted optical admittance of the "substrate" or outgoing medium, whereas [tex]\eta_0[/tex] corresponds to the incident medium. They can be made the same, but are not required to always be the same. That's the point. With that in mind, I should say expression (12) is a more appropriate description. Sounds simple, but I can be wrong. Please let me know what you think. Thanks!
     
    Last edited: Apr 14, 2008
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