# Optical coherence tomography

In Optical Coherence Tomography, Broader bandwidth/low coherence light sources are used. Is it only because to increase the resolution or are there any other reasons? What will happen if we use a high coherence source ?

Another way to put this: by scanning the reference arm length, you are measuring the light's autocorrelation function $R(\Delta x)$. By the Wiener-Khinchin theorem, the light's power spectral density as a function of wavenumber $k = \omega/c$ is the Fourier transform of $R(\Delta x)$. So, for a resolution $\Delta x$ of $1{\rm \mu m}$, we must have $\Delta k \approx 1{\rm (\mu m)^{-1}}$, and, given $\Delta k = 2\pi \Delta \lambda/\lambda^2$, this means a source of breadth $\Delta \lambda \approx 0.16 \mu m$, or approximately 160nm linewidth when $\lambda = 1000nm$.
It is wholly a question of imaging depth. The system is working like a confocal microscope (see my description here) and this is your answer for when high coherence light is used. However, to image deeply into samples, for example, the retina of an eye, one can generally only use very low numerical aperture, owing to the high wavefront aberration imparted by the eye ("high" in confocal microscope terms - we're still only talking a wave or a few waves of aberration) on the signal as it travels to and fro through the eye and retina. The higher the numerical aperture of a confocal microscope, the finer its depth resolution, but unfortunately the tolerance to wavefront aberration becomes swiftly less with bigger NA. To image deep into tissue, we are therefore restricted to numerical apertures of 0.1 or less, which translates to a depth resolution of roughly $100{\rm \mu m}$ to $200{\rm \mu m}$. The use of OCT restores the depth resolution in an otherwise poorly depth resolving confocal microscope system.