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Derivation of TDOCT axial resolution V2

The derivation is quite good in Chapter 9 of the book Biomedical Optics: Principles and Imaging [1] by Lihong V. Wang, so I’m not going to repeat again. Here I will add some notes which might be helpful during the derivation. The formula of coherence length is directly given without specific derivation (Eq. 9.15). {l_c} […]

The derivation is quite good in Chapter 9 of the book Biomedical Optics: Principles and Imaging [1] by Lihong V. Wang, so I’m not going to repeat again. Here I will add some notes which might be helpful during the derivation.


The formula of coherence length is directly given without specific derivation (Eq. 9.15).

{l_c} = \frac{{4\ln 2}}{\pi }\frac{{\lambda _0^2}}{{\Delta \lambda }}

I give an approximate derivation here, and the notation is in accordance with that in the book.

\left\{ \begin{array}{l}
\nu {\rm{ = }}\frac{c}{{{\lambda _0}}}\\
\Delta \nu  = \frac{{c \cdot \Delta \lambda }}{{\lambda _0^2}}\\
{l_c} = c{\tau _c} = \frac{c}{{\Delta \nu }} = \frac{{\lambda _0^2}}{{\Delta \lambda }}
\end{array} \right.

where \nu denotes the frequency of light source, and \Delta \nu denotes the FWHM bandwidth in frequency.

Some coefficients are neglected here. For more specific details, please read the paper Estimation of longitudinal resolution in optical coherence imaging [2].

To prove the formula,

{l_c} = 8\ln 2 \cdot \frac{c}{{\Delta \omega }}

we know that \omega = 2\pi \nu, so it is intuitive to get \Delta \omega = 2\pi \cdot \Delta \nu.

\Delta \omega  = 2\pi  \cdot \Delta \nu  = \frac{{2\pi c \cdot \Delta \lambda }}{{\lambda _0^2}}

For Eq. (9.21), the real number term,

2{\mathop{\rm Re}\nolimits} \left\{ {{E_R}\left( \omega  \right)E_S^*\left( \omega  \right)} \right\} = {\mathop{\rm Re}\nolimits} \left\{ {{E_R}\left( \omega  \right)E_S^*\left( \omega  \right)} \right\} + {\mathop{\rm Re}\nolimits} \left\{ {{E_S}\left( \omega  \right)E_R^*\left( \omega  \right)} \right\}

Pay attention to introduction of the round-trip phase delay \Delta {\tau _p}, the round-trip group delay \Delta {\tau _g}, phase velocity v_p and group velocity v_g,

\left\{ \begin{array}{l}
{v_p} = \frac{{{\omega _0}}}{{k\left( {{\omega _0}} \right)}}\\
\Delta {\tau _p} = \frac{{k\left( {{\omega _0}} \right)}}{{{\omega _0}}}\left( {2\Delta l} \right) = \frac{{2\Delta l}}{{{v_p}}}\\
{v_g} = \frac{1}{{k'\left( {{\omega _0}} \right)}}\\
\Delta {\tau _g} = k'\left( {{\omega _0}} \right)\left( {2\Delta l} \right) = \frac{{2\Delta l}}{{{v_g}}}
\end{array} \right.

Also, think about this expression: If S\left( \omega \right) is symmetric about {\omega _0}, the integral in Eq. (9.36) is real.

Why is it? You can consider the property of the even function.

After that, Eq. (9.37) can be obtained,

{I_{{\rm{AC}}}} \propto \cos \left( {{\omega _0}\Delta {\tau _p}} \right)\int_{ - \infty }^\infty  {S\left( \omega  \right)\exp \left( { - i\left( {\omega  - {\omega _0}} \right)\Delta {\tau _g}} \right)d\omega } \label{int-g}

Further, if S\left( \omega \right) is Gaussian, in case that you forget, that is,

S(\omega)=\frac{1}{\sqrt{2 \pi} \sigma_{\omega}} \exp \left(-\frac{\left(\omega-\omega_{0}\right)^{2}}{2 \sigma_{\omega}^{2}}\right)

Therefore,

\begin{array}{l}
{I_{{\rm{AC}}}} \propto \cos \left( {{\omega _0}\Delta {\tau _p}} \right)\int_{ - \infty }^\infty  {\frac{1}{{\sqrt {2\pi } {\sigma _\omega }}}\exp \left( { - \frac{{{{\left( {\omega  - {\omega _0}} \right)}^2}}}{{2\sigma _\omega ^2}}} \right)\exp \left( { - i\left( {\omega  - {\omega _0}} \right)\Delta {\tau _g}} \right)d\omega } \\
 = \frac{{\cos \left( {{\omega _0}\Delta {\tau _p}} \right)}}{{\sqrt {2\pi } {\sigma _\omega }}}\int_{ - \infty }^\infty  {\exp \left[ {\frac{{ - {\omega ^2} - \left( { - 2{\omega _0} + i2\sigma _\omega ^2\Delta {\tau _g}} \right)\omega  - \left( {\omega _0^2 - i2\sigma _\omega ^2\Delta {\tau _g}{\omega _0}} \right)}}{{2\sigma _\omega ^2}}} \right]d\omega } 
\end{array}

Here, we refer to the general form of Gaussian integral [3]:

\int_{ - \infty }^\infty  {\exp \left( { - a{x^2} - bx - c} \right)dx}  = \sqrt {\frac{\pi }{a}} \exp \left( {\frac{{{b^2}}}{{4a}} - c} \right)

We have,

\left\{ \begin{array}{l}
a = \frac{1}{{2\sigma _\omega ^2}}\\
b = \frac{{ - 2{\omega _0} + i2\sigma _\omega ^2\Delta {\tau _g}}}{{2\sigma _\omega ^2}}\\
c = \frac{{\omega _0^2 - i2\sigma _\omega ^2\Delta {\tau _g}{\omega _0}}}{{2\sigma _\omega ^2}}
\end{array} \right.

Next, we do a series of calculation,

\left\{ \begin{array}{l}
{b^2} = \frac{{ - 4\sigma _\omega ^4\Delta \tau _g^2 - i8\sigma _\omega ^2\Delta {\tau _g}{\omega _0} + 4\omega _0^2}}{{4\sigma _\omega ^4}}\\
4a = \frac{2}{{\sigma _\omega ^2}}\\
\frac{{{b^2}}}{{4a}} = \frac{{ - 4\sigma _\omega ^4\Delta \tau _g^2 - i8\sigma _\omega ^2\Delta {\tau _g}{\omega _0} + 4\omega _0^2}}{{8\sigma _\omega ^2}}\\
c = \frac{{4\omega _0^2 - i8\sigma _\omega ^2\Delta {\tau _g}{\omega _0}}}{{8\sigma _\omega ^2}}\\
\frac{{{b^2}}}{{4a}} - c = \frac{{ - \sigma _\omega ^2\Delta \tau _g^2}}{2}\\
\sqrt {\frac{\pi }{a}}  = \sqrt {2\pi } {\sigma _\omega }
\end{array} \right.

Finally, I_{AC} is obtained as below,

{I_{{\rm{AC}}}} \propto \cos \left( {{\omega _0}\Delta {\tau _p}} \right)\exp \left( {\frac{{ - \sigma _\omega ^2\Delta \tau _g^2}}{2}} \right)

It is actually the Eq. (9.38) in the book.


Let’s have a look at Eq (9.41), for the spectrum with a normal distribution, the relationship between its FWHM and the standard deviation is [4],

\Delta \xi  = 2\sqrt {2\ln 2} {\sigma _\xi }

Finally, the axial resolution of OCT in air \Delta {z_R} is obtained,

\Delta {z_R} = \frac{{2\ln 2}}{\pi }\frac{{\lambda _0^2}}{{\Delta \lambda }}

Therefore, the axial resolution in air equals half of the coherence length of the source owing to the round-trip propagation of the reference and the sample beams [1].


The transverse resolution of OCT in air \Delta {r_R} is also given,

\Delta {r_R} = \frac{{4{\lambda _0}}}{\pi }\frac{f}{D} \approx \frac{{2{\lambda _0}}}{\pi }\frac{1}{{{\rm{NA}}}}

In addition, the depth of focus \Delta {z_f}, which is twice of the Rayleigh range, is given as below,

\Delta {z_f} = \frac{{\pi \Delta r_R^2}}{{2{\lambda _0}}}

Reference

[1] Biomedical Optics: Principles and Imaging

[2] Estimation of longitudinal resolution in optical coherence imaging

[3] Gaussian integral

[4] FWHM

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