[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

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

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!

 

[astrokreng]

Hi there! I would like to commend you for your thorough understanding and application of Macleod's expression for phase shift upon reflection. Your derivation is correct and well-explained.

To answer your question, the phase shift upon reflection is an important phenomenon in thin film optics. It is caused by the difference in optical admittance between the incident and reflected waves, which is determined by the characteristic matrix of the thin film.

In simple terms, the phase shift occurs because the thin film acts as a waveguide for the reflected wave, causing it to travel a longer path than the incident wave. This results in a phase difference between the two waves, which can be calculated using the reflection coefficient (ρ) as shown in your derivation.

I hope this helps to clarify your understanding of phase shift upon reflection in thin films. Keep up the good work in your studies!