Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Lets say I want to design a Nomarski prism that would split the ordinary and extraordinary beam by an angle of 0.32 mrad. I used a raytracer to find the internal angle between the quartz wedges.

However I couldn't decide how thick the quartz wedges should be.

The original patent mentions the thickness but I don't understand the sentence:

Picture of Nomarski Patent

The picture shows the Nomarski prism from the side. The optical axis of the system is horizontal. The internal angle between the wedges (or blades) is $\alpha$. The optic axis (note the difference between optical axis of the system and optic axis of the crystal) of the first prism Q2 is perpendicular to the screen. The wedge Q1 has its axis inclined in-plane.

Q2 is 0.5mm thick at the center. Q1 is 1mm thick at the center.

Both compensate their birefringence.

This last expression isn't clear to me. I have the feeling that I understand this only vaguely.

  1. Is the birefringence in Q2 stronger because of the orientation of the optic axis? (Probably yes)

  2. Why does the birefringence need to be compensated? (must have something to do with the adjusting the bias by shifting the Nomarski prism transversal to the optical axis in the microscope)

  3. How is the total thickness of the prism (here 1.5mm) chosen? Can a thinner Nomarski prism be used with higher angles?

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.