# What are some formulas that are associated with the Schmidt corrector?

I want to fully understand how an SCT works. Thus, I figured out I can use some formulas that describe how an incident ray that is parallel with the axis would act after refracting off a Schmidt plate. I presume that the formulas would depend on the vertical and horizontal distances from the center of the plate. My goal now is to make a ray tracing model of an SCT with certain specifications. Please make your answer as simple as possible.I will be thankful if a nontechnical source is provided at the end of the answer.

• I know this formula for the density of a Schmidt corrector: $\rho_{Schmidt} = M_{Schmidt} / V_{Schmidt}$ but this may not be very useful ;-) – uhoh Jan 6 '19 at 1:54
• I can't thank you enough... This changed my life! – Thamer AL Sadoun Jan 8 '19 at 18:03

The specific purpose of a Schmidt Corrector plate is to have an equal but opposite spherical aberration to the primary mirror they compensate for. So any formula you try to come up with will depend on the primary.

In fact the information on SCT's shows how it is made using physical forms, not using formulae.

From revolvy.com:

A thin glass disk with a perfectly polished accurate flat form was placed on a heavy metal pan. The upper edge of the pan was ground at a precise angle or bevel based on the coefficient of elasticity of the particular type of glass plate that was being used. The glass plate was sealed to the ground edge of the pan, then a vacuum pump was used to exhaust the air until a particular negative pressure had been achieved. This caused the glass plate to warp slightly. The exposed side was then ground and polished to a perfect flat. When the vacuum was released, the plate sprang back until its bottom surface was again plane, while the upper surface had the correct figure.

If, however, you do wish to work out the SCT parameters from formulae, this page on telescope-optics.net has the fairly complex info you need to use this formula:

$$z=\frac{(\rho^4-\Lambda\rho^2)d^4}{4(n^\prime-n)R^3}=\frac{(\rho^4-\Lambda\rho^2)D}{512(n^\prime-n)F^3}$$

It's not an easy page to work through - but for someone who hasn't needed complicated maths for some years I could just about handle it.

• I may have put the non-technical answer at the start, and ended with the technical one... :-) – Rory Alsop Jan 3 '19 at 20:05
• Please use Mathjax for mathematics like this, rather than posting an image. It's the site policy (pretty common to all science SEs). – StephenG Jan 4 '19 at 4:34
• @StephenG - I finally managed to convert it to MathJax - that's not easy at all! – Rory Alsop Jan 6 '19 at 9:22
• It get's easier with practice. I made a minor correction to use the prime symbol rather than the '1' you used. The prime symbol is a little obscure starting off with Mathjax (which is almost the same as Latex). – StephenG Jan 6 '19 at 9:40
• Thanks Stephen - I've favourited 2 mathjax pages so I can muddle through next time. – Rory Alsop Jan 6 '19 at 9:49