## Abstract

Mirror misalignment or the tilt angle of the Michelson interferometer can be estimated from the modulation depth measured with collimated monochromatic light. The intensity of the light beam is usually assumed to be uniform, but, for example, with gas lasers it generally has a Gaussian distribution, which makes the modulation depth less sensitive to the tilt angle. With this assumption, the tilt angle may be underestimated by about 50%. We have derived a mathematical model for modulation depth with a circular aperture and Gaussian beam. The model reduces the error of the tilt angle estimate to below 1%. The results of the model have been verified experimentally.

© 2011 Optical Society of America

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### Equations (15)

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(1)
$${I}_{B}(x)={\int}_{-2\alpha R}^{2\alpha R}I(x+y)2\sqrt{{(2\alpha R)}^{2}-{y}^{2}}{[\pi {(2\alpha R)}^{2}]}^{-1}\mathrm{d}y.$$
(2)
$$B(y)=\{\begin{array}{ll}2\sqrt{{(2\alpha R)}^{2}-{y}^{2}}{[\pi {(2\alpha R)}^{2}]}^{-1}& |y|\le 2\alpha R\\ 0& \text{elsewhere}.\end{array}$$
(3)
$$I(x)={\int}_{-\infty}^{\infty}E(\nu )\mathrm{exp}(i2\pi \nu x)\mathrm{d}\nu =\mathcal{F}\{E(\nu )\},$$
(4)
$${I}_{B}(x)={\int}_{-\infty}^{\infty}E(\nu )\mathrm{exp}(i2\pi \nu x)[{\int}_{-\infty}^{\infty}B(y)\mathrm{exp}(i2\pi \nu y)\mathrm{d}y]\mathrm{d}\nu .$$
(5)
$${m}_{B}=\frac{2{J}_{1}(4\pi \nu R\alpha )}{4\pi \nu R\alpha}=\frac{2{J}_{1}(\pi k)}{\pi k},$$
(6)
$$g(t,u)={g}_{0}\text{\hspace{0.17em}}\mathrm{exp}(-a{t}^{2}-a{u}^{2}),a={(2{\sigma}^{2})}^{-1},$$
(7)
$$G(y)\frac{\mathrm{d}y}{2\alpha}=\mathrm{exp}[-\frac{a{y}^{2}}{{(2\alpha )}^{2}}]\left[{\int}_{-q(y)}^{q(y)}\text{\hspace{0.17em}}\mathrm{exp}(-a{u}^{2})\mathrm{d}u\right]\frac{\mathrm{d}y}{2\alpha},$$
(8)
$${I}_{G}(x)={\int}_{-\infty}^{\infty}E(\nu )\mathrm{exp}(i2\pi \nu x)\phantom{\rule{0ex}{0ex}}\times [{G}_{0}{\int}_{-\infty}^{\infty}G(y)\mathrm{exp}(i2\pi \nu y)\mathrm{d}y]\mathrm{d}\nu \mathrm{.}$$
(9)
$${m}_{G}={G}_{0}{\int}_{-2\alpha R}^{2\alpha R}\text{\hspace{0.17em}}\mathrm{exp}[-\frac{a{y}^{2}}{(2\alpha {)}^{2}}]\phantom{\rule{0ex}{0ex}}\times \left[{\int}_{-q(y)}^{q(y)}\text{\hspace{0.17em}}\mathrm{exp}(-a{u}^{2})\mathrm{d}u\right]\mathrm{exp}(i2\pi \nu y)\mathrm{d}y.$$
(10)
$${\int}_{-q(y)}^{q(y)}\mathrm{d}u=2q(y)=2\pi \alpha {R}^{2}B(y),$$
(11)
$${m}_{G}\approx {m}_{G*}={M}_{0}{\int}_{-\infty}^{\infty}B(y)\mathrm{exp}[-\frac{a{y}^{2}}{(2\alpha {)}^{2}}]\mathrm{exp}(i2\pi \nu y)\mathrm{d}y,$$
(12)
$${m}_{G*}={M}_{0}(\mathcal{F}\{B(y)\}*\mathcal{F}\left\{\mathrm{exp}[-\frac{{y}^{2}}{2{(2\alpha \sigma )}^{2}}]\right\}),$$
(13)
$$\mathcal{F}\left\{\mathrm{exp}[-\frac{{y}^{2}}{2(2\alpha \sigma {)}^{2}}]\right\}=2\sqrt{\frac{2}{\pi}}\sigma \alpha \text{\hspace{0.17em}}\mathrm{exp}[-2{(2\pi \sigma \alpha \nu )}^{2}].$$
(14)
$$c\left(\frac{\sigma}{2R}\right)=1.23-\frac{0.23}{1+{[4.3\sigma /(2R)]}^{2.8}}.$$
(15)
$${m}_{Gc}(k,\sigma )={m}_{G*}(k,{\sigma}_{c})={m}_{G*}(k,\sigma c).$$