So as per my understanding the pixel size (let's say in microns) of a camera like ccd or scmos would also depend on the rest of the optical setup like the objective used. So one would carry out a calibration experiment like with a nano-positioning stage to calculate the pixel to micron ratio. But I found this website ( https://bostonmicroscopes.com/product/andor-zyla-5-5-scmos-monochrome-microscope-camera/ ) where they have mentioned the pixel size as part of the camera specs. How is it possible? Also in another paper the authors mentioned the imaging area (say, x micron sq.) and then ROI (say y x y pixels) and so mentioned the pixel size as x/y. So how do I get this imaging area?
Aparna Sathya Murthy I think I got the physical pixel size of the camera confused with the projected pixel size, may be? Like there would be a pixel to micron conversion factor for my image right?
What is mentioned in the specification (6.5 μm) is actually the size of a single CCD sensor and it is calculated by dividing the total width of the sensor array (16.6 mm) by the total number of pixels generated (2560) i.e. 16.6 mm / 2560 = 6.5 μm . To actually calculate the pixel size of an image the resolution in dpi should be known e.g. if the camera supports an optical resolution (not interpolated resolution) of say 1200 dpi, then the pixel size is (1/1200) inch.
Arti Tyagi : the term "pixel size" is used with different meanings.
One usage is the physical size of the sensitive region of an individual CCD or CMOS sensor element. Sometimes referred to as "sensel size" to distinguish it from pixel pitch. Note that many sensors include micro-lenses which increase the effective size of the sensor element, compared with the physical dimensions underlying sensitive area. This effective size is relevant to understanding the precise resolution and system transfer function of your apparatus.
For your purposes, the pixel pitch is more important, and is the distance between adjacent pixels in the sensor array. While often referred to as "pixel size", this is usually larger than the size of the sensitive elements, and as Aparna Sathya Murthy points out, is fixed with high precision during manufacture of the sensor.
It seems you wish to calibrate the dimensions of your images, so you need to take into account the magnification of your optical system. The most straightforward approach is to image a subject of known size, such as a calibration test target:
https://www.thorlabs.com/newgrouppage9.cfm?objectgroup_id=7502
https://www.edmundoptics.co.uk/c/test-targets/1099/#
https://www.edmundoptics.co.uk/f/precision-ronchi-rulings-on-opal-glass/13366/
If your configuration allows it, you could instead use a calibrated movement of your subject, as you suggest in your original question. You may be able to use a dial gauge, interferometer, or some other displacement sensor in combination with an un-calibrated stage.
The most suitable method will depend on your specific configuration, how precise your calibration must be, the equipment you have available or can borrow, and the size of your equipment budget.
Understood, thank you Alan Robinson , Aparna Sathya Murthy and Ranjan Parekh !
Hi Arti, 6.5um x 6.5um is a good size for a pixel on a microscope. To avoid an effect called aliasing you need two pixels across to view a 13um line or feature at the sensor in the camera (which is about as fine a line as most microscopes can draw no matter what the magnification). You run into problems with diffraction as you try for higher magnifications - the fine detail in the image fades away as the lens system is unable to resolve it. 6.5um is a good match the Numerical aperture of most microscope objectives. But basically the other contributors are correct the sensor pixel is fixed at the silicon foundry. The variable you are looking to measure is the magnification in your optics.
To calculate the pixel size from your camera system (that includes the camera in question) use these equations that are proven and work: Pixel = EFL*lambda/(Q*D); EFL is the effective focal length, lambda is center wavelength, D is the effective aperture and Q is the quality factor. Q is often set to 2.0, which is the Nyquist. I choose to know Q by measurement. Input a low power laser into your aperture (use much NDs) or use a pinhole (white light) and image the PSF at the focus. Then zoom in on the PSF image and count the number of pixels across the PSF 1/e2 diameter. Now use this equation to find Q. Q = (EFL*lambda*# of pixels)/PSF. You must know your EFL, the effective diameter of your input light and the center wavelength. PSF = 2.44*lambda*f/#. Hope this helps.
Logan DesAutels so in that psf equation, f is the focal length and # is the number of pixels in the 1/e2 diameter of the psf image?
Arti Tyagi: the equation for PSF given in Logan DesAutels's answer looks like that for the diameter of the first dark ring of the Airy disk in a diffraction-limited optical system. https://en.wikipedia.org/wiki/Airy_disk
PSF_Diameter = 2.44 λ N, where λ is the wavelength and N is the normalised aperture or f-number of the system.
N = f / D is the ratio of effective focal length, f, to the entrance pupil diameter, D, of the lens. It is sometimes written (f/#) or (F#).
F-number is more often used to describe telescopes or photographic lenses than in microscopy. You may prefer to use the relationship
N = 1 / (2 NA) where NA is the numerical aperture of the optical system.
https://en.wikipedia.org/wiki/Numerical_aperture#Numerical_aperture_versus_f-number
Note that PSF_diameter = 2.44 λ N is NOT the 1/e2 diameter of the point spread function, unless you have a rather large and very specific amount of aberration in your system. The 1/e2 diameter of a diffraction-limited PSF is much less than this.
https://en.wikipedia.org/wiki/Airy_disk#Approximation_using_a_Gaussian_profile
If you use the numerical aperture of your microscope objective lens, this gives the object space PSF diameter (at your subject). To convert to a PSF at your image sensor, multiply by the optical magnification, or use the image-space numerical aperture if this is known.
In Logan DesAutels's first equation, w = ( f λ )/( Q D ) = N λ / Q, the maximum spatial frequency in a captured image is 1/(N λ), so if w is the pixel pitch, Q is the number of samples within a single period of the maximum spatial frequency. In principle (ignoring noise and other real-world problems) for sample rates higher than Nyquist (Q > 2) there is sufficient information to reconstruct the original optical intensity distribution in the image, without sampling artefacts such as aliasing.
https://spie.org/publications/tt52_151_diffraction_mtf?SSO=1
https://en.wikipedia.org/wiki/Nyquist_rate
https://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem
If you are only interested in distance scaling of your images, this does not matter. However, sampling theory and Q are valuable concepts when deciding which magnification to use, and understanding how much detail it might be possible to recover.
Yes, sorry, to be consistent with my other equation variables, the PSF or Airy 1st minimum diameter is defined as PSF = 2.44*lambda*EFL/D.
Do you know any of the parameters I mentioned like EFL, effective D, etc... and are you in a laboratory setting? If not then you could try a simpler experiment to measure the PSF, which is to just place a bright light behind a very small hole at the farthest distance you can get in your home. You can buy a pinhole fairly cheap at Thorlabs or Edmund Optics or you can make one with tin foil and a needle. For white light use 587nm as your center wavelength. You could place a cheap laser diode (purchase online for $20) behind the pinhole and use its wavelength). Then take an image of this pinhole and you should be able to observe the PSF or Airy rings. Count the pixels across the center spot within the 1st Airy minimum and that is the number of pixels for the Q equation. You will have to know your lens EFL and D or if it is a photography lens use the f# (f-number) setting because F# = EFL/D. Exposure time does not matter; just get the best image you can. You can use any F# too, but you just need to know what it is. Probably the smaller F# the better, but use whatever gives you the best image.
Logan DesAutels thank you for your suggestion, I understand these parameters and my questions were for a lab experiment.
Logan DesAutels I need to measure the size of my detection volume. now in my case i know the axial diameter but i need to know the lateral diameter. It is a TIRF based system but the illumination setup is a little different, basically illuminated from one end of the sample containing glass slide. So here the lateral dimension of the detection volume will be defined by the objective right? will it be what is called the 'beam waist' like the diameter of light beam collected by the objective? and how do I find that?
Arti I'm sorry I'm not understanding what you are trying to describe. Are talking about measuring the x and y pixel dimensions of your sensor? TIRF? Sorry I couldn't help.
Logan DesAutels I meant total internal reflection fluorescence microscopy. Sorry my description was not clear, but I understood I was just looking for a formula to calculate the focused beam spot size coming from an objective lens. So I got my answer!
https://greenfluorescentblog.wordpress.com/2013/01/10/calculating-the-pixel-sizes-on-images/
I hope, you will find something in this.
https://static1.squarespace.com/static/551ecf1ae4b0b101cf72bfa3/t/55412684e4b0512f43caa5de/1430333060072/Resolution_calculation.pdf
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