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

  • #1
372
0

Main Question or Discussion Point

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)
 

Answers and Replies

  • #2
372
0
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:

Related Threads for: [thin films] (Macleod's) phase-shift upon reflection

Replies
1
Views
3K
  • Last Post
Replies
4
Views
6K
Replies
4
Views
763
Replies
1
Views
2K
Replies
1
Views
941
Replies
4
Views
502
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
5
Views
2K
Top