I could use last week an optical microscope, didn't seem special in any way, 50x magnification, image viewable per a CCD camera on a computer screen besides through the ocular.

But the software of this microscope had a feature to show a 3D calculated topography view of the sample (silicon wafer with some laterally structured micron sized 40 nanometer high SiO2 squares). So in the software you could see the height of these squares (30-50 nm there, so some error) and also do virtual profilometer over these squares. There these height measurements were separated for green, red, blue, so 3 lines showing a 30-40 nm difference from silicon to a SiO2 square.

The manual didn't tell how the physics work, so can has someone here an idea how this might work. It looked like a cheap normal microscope, I suppose that it uses somehow the CCD data and reflected intensity of the squares, the height of the bigger squares was more accurately measured/extrapolated than smaller squares.

Also I can exclude it is a Normaski microscope


there were no polarisators or prisms inside the microscope.


This technique seems to come near, white light interferometry, but resolution is much better.

So can someone explain to me how this technique that is probably used in this microscope is called and what the accuracy is, what limits it?

Link to the specific model https://www.micro-shop.zeiss.com/?s=12128207d6c5cf&l=en&p=us&f=e&i=10247&o=&h=25&n=0&sd=490950-0002-000#490950-0002-000

  • 1
    $\begingroup$ There are many ways to do this. It's going to be necessary to know exactly what type of microscope you are using. If you don't know the type, you should be able to Google the model name. Or at least tell us the model name. $\endgroup$
    – Colin K
    Commented Mar 11, 2013 at 22:59
  • $\begingroup$ @ColinK added link above... maybe explanation phys.org/news/… $\endgroup$
    – James Last
    Commented Mar 12, 2013 at 13:46

1 Answer 1


By measuring the z -position of the stage at focus, the height of objects can be determined.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.