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

[DivGradCurl]

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!

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

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

\rm \eta_0 E_i - \eta_0 E_r = \eta_1 E_t \qquad (3)

Substitute (1) into (3):

\rm \eta_0 E_i - \eta_0 E_r = \eta_1 \left( E_i + E_r \right) \qquad (4)

Rearrange (4):

\rm E_i \left( \eta_0 - \eta_1 \right) = E_r \left( \eta_0 + \eta_1 \right) \qquad (5)

Apply the definition of reflection coefficient \rho:

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

Replace \eta_1 by the input optical admittance Y in (6):

\rho = \frac{ \eta_0 - Y}{ \eta_0 + Y} \qquad (7)

By definition, Y \equiv C/B. Then

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

From the general form of the characteristic matrix

\begin{bmatrix} <br /> \displaystyle B \\<br /> \displaystyle C <br /> \end{bmatrix} = <br /> \begin{bmatrix}<br /> \displaystyle \alpha &amp; \displaystyle i\beta \\<br /> \displaystyle i\gamma &amp; \displaystyle \delta <br /> \end{bmatrix} <br /> \begin{bmatrix}<br /> \displaystyle 1 \\<br /> \displaystyle \eta_m<br /> \end{bmatrix} =<br /> \begin{bmatrix}<br /> \displaystyle \alpha + i\beta \eta _m \\<br /> \displaystyle \delta \eta_m + i \gamma<br /> \end{bmatrix} \qquad (9)<br />

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

(problems previewing all of this latex code in one post - I'll break it up - please wait)

 

[DivGradCurl]

OK. Part II...

Therefore

\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)

so

\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)

By inspection of (11), it follows

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

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

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

Note: \eta_m is the tilted optical admittance of the "substrate" or outgoing medium, whereas \eta_0 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!