## Abstract

We derive a simple equation to predict the center-wavelength shift of a Fabry-Perot type narrow-bandpass filter by using the conventional characteristic matrix method and the elastic strain model as the temperature varies. We determine the thermal expansion coefficient of substrate from the zero-shift condition of the center wavelength of the filter. The calculated shifts are in a good agreement with the experimental ones, in which the narrow-bandpass filters are prepared by plasma ion-assisted deposition on four substrates with different thermal expansion coefficients.

© 2004 Optical Society of America

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### Equations (18)

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(1)
$${n}_{f}^{T}={n}_{f}^{0}+\left[{\left(\frac{dn}{dT}\right)}_{f}\Delta T+\left(1-{n}_{f}^{0}-{\left(\frac{dn}{dT}\right)}_{f}\Delta T\right)\left(\frac{{A}_{f}\Delta T}{1+\left(3{\alpha}_{f}+{A}_{f}\right)\Delta T}\right)\right]$$
(2)
$${d}_{f}^{T}={d}_{f}^{0}\left\{1+\left({\alpha}_{f}-{B}_{f}\right)\Delta T\right\},$$
(3)
$${A}_{f}=\frac{2\left(1-2{v}_{f}\right)}{\left(1-{v}_{f}\right)}\left({\alpha}_{s}-{\alpha}_{f}\right)\phantom{\rule{.5em}{0ex}}\mathrm{and}\phantom{\rule{.5em}{0ex}}{B}_{f}=\frac{2{v}_{f}}{\left(1-{v}_{f}\right)}\left({\alpha}_{s}-{\alpha}_{f}\right),$$
(4)
$$\left[{M}_{f}\right]=\left[\begin{array}{cc}\mathrm{cos}{\delta}_{f}& \left(i\mathrm{sin}{\delta}_{f}\right)\u2044{n}_{f}\\ i{n}_{f}\mathrm{sin}{\delta}_{f}& \mathrm{cos}{\delta}_{f}\end{array}\right]$$
(5)
$${\delta}_{f}=\frac{2\pi}{\lambda}{n}_{f}{d}_{f}.$$
(6)
$${\delta}_{f}^{T}={\delta}_{f}^{0}+{\epsilon}_{f},$$
(7)
$${\left[M\right]}^{T}=\left[\begin{array}{cc}{M}_{11}& {M}_{12}\\ {M}_{21}& {M}_{22}\end{array}\right]={\left(\left[{M}_{C}\right]\left[{M}_{D}\right]\right)}^{p}{\left(2\left[{M}_{C}\right]\right)}^{q}{\left(\left[{M}_{D}\right]\left[{M}_{C}\right]\right)}^{p},$$
(8)
$${Y}^{T}=\frac{{C}^{T}}{{B}^{T}},$$
(9)
$${\left[\begin{array}{c}B\\ C\end{array}\right]}^{T}={\left[M\right]}^{T}\left[\begin{array}{c}1\\ {n}_{S}^{T}\end{array}\right],$$
(10)
$${R}^{T}={\mid \frac{{n}_{0}^{T}-{Y}^{T}}{{n}_{0}^{T}+{Y}^{T}}\mid}^{2},$$
(11)
$${R}_{min}^{T}={\mid \frac{{n}_{0}^{T}-{n}_{s}^{T}}{{n}_{0}^{T}+{n}_{s}^{T}}\mid}^{2},$$
(12)
$${Y}^{T}={n}_{S}^{T}.$$
(13)
$$P{\epsilon}_{H}+Q{\epsilon}_{L}=0,$$
(14)
$$P={n}_{H}^{T}{n}_{L}^{T}\phantom{\rule{.5em}{0ex}}\mathrm{and}\phantom{\rule{.5em}{0ex}}Q={\left({n}_{L}^{T}\right)}^{2}+\left\{{\left({n}_{H}^{T}\right)}^{2}-{\left({n}_{L}^{T}\right)}^{2}\right\}q$$
(15)
$$P={\left({n}_{L}^{T}\right)}^{2}+\left\{{\left({n}_{H}^{T}\right)}^{2}-{\left({n}_{L}^{T}\right)}^{2}\right\}q\phantom{\rule{.4em}{0ex}}\mathrm{and}\phantom{\rule{.4em}{0ex}}Q={n}_{H}^{T}{n}_{L}^{T}$$
(16)
$${\epsilon}_{f}=\frac{2\pi}{{\lambda}_{T}}{n}_{f}^{T}{d}_{f}^{T}-\frac{\pi}{2}$$
(17)
$${\lambda}_{T}=4\left[\frac{P\left({n}_{H}^{T}{d}_{H}^{T}\right)+Q\left({n}_{L}^{T}{d}_{L}^{T}\right)}{P+Q}\right]$$
(18)
$$\frac{\Delta \lambda}{{\lambda}_{0}}=\frac{{\lambda}_{T}-{\lambda}_{0}}{{\lambda}_{0}}=\frac{4}{{\lambda}_{0}}\left[\frac{P\left({n}_{H}^{T}{d}_{H}^{T}\right)+Q\left({n}_{L}^{T}{d}_{L}^{T}\right)}{P+Q}\right]-1$$