I have been struggling with this concept for a very long time. To calculate the total magnification provided by any digital microscope we generally multiply all the individual magnifications (objective, ocular, coupler/adapter, and the video/on-screen magnification). But currently, I'm working on an in-house developed system and I'm trying to match the magnification to that of a traditional microscope so when I modify the distance between the sensor and the objective lens, my total magnification obviously changes. How do I calculate the total magnification here since the above formula isn't suitable for this case?
ps. not using any ocular here, a coupler lens is also optional.
The magnification values printed on objective lenses do describe the actual magnification from the object to an intermediate image plane. In a modern infinity corrected microscope a tube lens is employed. In that kind of microscope the objective images to infinity and the tube lens, with a focal length of, say, 200mm, creates a real image close before the ocular or directly on a camera chip. The magnification M of that assembly equals the focal length ratio: M=f_tube/f_objective, this is the value that is imprinted on the objective. The objective alone has in fact no quantifyable magnification. In a finite image distance microscope, like you descibe it, when you image directly from the objective onto a camera chip, then the magnification is the ratio: M=image_distance/object_distance. If your image distance (maybe 160mm) is significantly larger than the focal length of the objective, then the object distance is about the focal length, and magnification becomes M=camera_distance_from_objective/f.
Rather than trying to calculate the magnification you could take an image of a stage micrometer and measure it.
Alexander Dräbenstedt that makes sense, but the value I get this way doesn't look right. The images are way more magnified. Is there some factor I'm missing here? And I'm sorry I forgot to mention it's a finite system (like you said camera mounted directly onto a fixed tube length objective).
Matthew J Gratian I did do that. actually. So I just get the digital pixel size (1px = so many microns) and compare these values to what I get from a commercial system?
Yes, exactly that. You can draw a scale bar on the image and that will be comparable between images from different instruments.
Matthew J Gratian Thank you! This is how I did it earlier but I thought there's an empirical way of getting the magnification so just for validation purposes, I started looking.
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