optics - Confocal Microscopy

In the context of Confocal Microscopy literature state, "spatial rejection of out of phase light".Is that mean only light which is pass through the pinhole is used and the rest is blocked ?

Answer

Informally you are right: the pinhole rejects out-of-plane information by only registering light through the pinhole.

Confocal Microscopy uses a focussed beam to light the sample as opposed to the uniform lighting of brightfield microscopy. An imaging optic is then focussed onto the same point as where the illumination system is focussed, and registers the reflexion / fluorescence from that point. Because the light is focussed, the two focusses must be scanned together over a plane.

Often in benchtop systems, the focussing beam is made by lighting through a pinhole. The light accepted by the imaging optic is also restricted, by a pinhole in a benchtop system. In the kind of systems that I have designed, a scanning single mode optical fibre replaces both pinholes: the fibre's single mode works as a sending and receiving antenna and has the same directional properties as an antenna. The antenna overlap integral replaces the pinhole.

It is more instructive, and more worthwhile from a systems analysis standpoint, to ask this question in signal to noise terms. And the essence of confocal microscopy is revealed by answering the question:

"Why can't I put my hand under the objective of a powerful brightfield microscope, light my hand strongly (with, say, a fibre bundle lightsource) and see the cells in my hand by bringing the microscope into focus on my skin?"

So I draw a brightfield microscope below.

I've drawn the illumination light field in blue and the light scattered from an object in green (most often I design systems for fluorescence), but the argument works just as well in reflexion mode. The main point is that we have out collector lens system imaging the object onto, say, a CCD array and the aberration free field is such that the photon coupling amplitude from scatterer to a lone CCD pixel (or a pinhole - both work as spatial filters) varies as $\exp(i\,k\,R)/(1+ (i R/R_0))$, where $R$ is the distance of the object from the focus of the collector system. Here I've drawn the object we want to see at the focus. But there are also out-of-focus scatterers lit just as brightly as the object and the formula $\exp(i\,k\,R)/(1+ (i R/R_0))$ means that, in the farfield, each scatterer contributes a power proportional to $1/R^2$ to the pixel. So far so good: it seems the sensitivity of the instrument to out-of-focus information drops off very swiftly with distance from focus. Not swiftly enough, for we must sum up the noise contribution from the whole out of focus volume. If we assume scatterers (noise objects) are roughly equally responsive and are uniformly distributed in the large, then we can do this summation in spherical shells as shown. So each scatterer in the spherical shell a distance $R$ from the focus contributes noise power proportional to $1/R^2$ but each spherical shell volume varies like $R^2$ so that roughly each spherical shell of thickness $\delta R$, no matter how far from focus, contributes a roughly equal noise power. Therefore, if the sample is infinitely thick, the noise power diverges. This is exactly the reason why one would expect a uniformly bright night sky in an infinitely long living, infinitely wide universe as in Olbers's paradox.

One cannot see the cells in one/s own hand with a brightfield microscope because the out-of-focus noise levels are divergent.

This is precisely the reason we must prepare microscope slides in a pathology laboratory: the divergent noise figure is controlled by physically "gating" the out-of-focus information: we simply slash it off with a knife!'

My drawing below repeats this calculation for confocal microscopy.

Here we structure the lighting by focusing it one pixel at a time on the point we want to image. A system I have worked with lights the sample with the image of the tip of a single mode optical fibre: this begets a convergent lightfield in the sample as shown in my drawing by the blue lightfield. The fluorescence (or reflected light) is also gathered by the single mode of the same optical fibre: the fibre tip thus replaces the one pixel in my drawing. The fibre's mode works as a sending and receiving antenna, with its directivity and gain reciprocally related as for any other antenna. The upshot of all this is now that, not only is there a $\exp(i\,k\,R)/(1+ (i R/R_0))$ coupling amplitude for light returning to the optical fibre's single mode, there is another $\exp(i\,k\,R)/(1+ (i R/R_0))$ excitation amplitude as well because the illumination field is focussed. Therefore, the probability that a noise scatterer is raised to a fluoresciung state varies like $1/R^2$ and, given that this happens, the probability that the photon couples back into the fibre also varies like $1/R^2$. This means that the noise power contributed by a spherical shell of radius $R$ centred on the focus and of thickness $\delta R$ is proportional to $R^2 \times R^{-2}\times R^{-2} \delta R = R^{-2} \delta R$. Therefore, the noise gathered from an infinite medium is now a finite number and the object tissue does not have to be sliced to be imaged by a confocal microscope.

So the confocal microscope doesn't magically reject all the out-of-focus information. It simply lessens the microscope's sensitivity to this noise such that the signal to noise ratio is manageable, rather than pathological as in the brightfield microscope trying to image unprepared, solid samples or as in Olbers's paradox.