andrew s

As Merlin 66 says it does. With the case of an just an objective you don't have any option but to refocus to account for the focal shift due to the filter. With a reducer/corrector you have two degrees of freedom. You can adjust the reducer/corrector distance and refocus. This allows you to bring the reducer/corrector to ccd distance back to the design distance and then refocus to account for the focal shift. In the case of the objective only it also cause a shift from the design optical distance and that is why a filter in a converging beam introduces aberrations. As light relief you might like to ponder the following. 101 optics says that for a simple converging lens rays parallel to the optic axis (and close enough to it) go through the focal point. Now why is this so? Well one way to explain it is that all the rays have the same optical path length. The ray through the center of the lens goes through the thickest part of the lens but takes the shortest distance through the air along the optic axis while a ray through the edge of the lens has the least glass to go through but the longest route through the air from the edge of the lens to the focal point. In fact you find t*n + r*1 = constant ( where t is the thickness of glass gone through with n it's refractive index and r the air distance with refractive index~1). If you think about it it also explains why a simple lens has a curved focal surface! This is an example of the Principle of least Action applied to optics, It led Feynman to his formulation of quantum mechanics QED the subject of another post! Regards Andrew PS I am sure you will be all pleased when I get some clear skies and so post here less! Got any free slots at your place Olly? PPS John posted while I was typing but I think I covered the point. I am now worried my diagram will cause you all to do the wrong thing.

I think I understand you now but don't agree with your conclusion. A corrector works by adding a specific optical path difference into the light path to bring the light rays to, say, a flat focal plane at a specific optical depth (distance) from the corrector. As air has a refractive index of ~1 the optical and geometric distances will be the same e.g. your 55mm. If you place another optical element after the corrector e.g. our 3mm filter and as its refractive index will be ~1.4 you no longer have the matching of optical and geometric distance so to get back to the correct optical distance of 55mm from the corrector to ccd you need the geometric distance to be ~ 54mm ( 51 mm of air and 3mm glass as 51*1 + 3*1.4 ~ 55). This effect was use in the past to design very fast F1 Schmidt cameras which were solid glass with a cutout for the film! Regards Andrew

Ken - I am not sure I fully understand your post but I would say that the design distance is most probably the optical depth (geometric distance * refractive index) so if you place a 3 mm filter in the optical path it adds to the optical depth so you need to reduce the geometric distance to compensate. I agree with you earlier post on this. The two issues I discussed above are not mechanical v optical but about two different questions you can ask both of which can have mechanical v optical distance issues. Regards Andrew

Hi John - I think the confusion here is because there are two different case and different definitions of back focus. Case one what does adding a filter do to the focal position if you add a filter without changing the geometric distances- in this case as my diagram show it moves the focal point further away from the objective i.e you add ~ 1/3 thickness to get the focal point. Case two what to do if you place a filter between a corrector and ccd. In this case you what to maintain the optical depth between corrector and ccd and you need to subtract ~ 1/3 thickness (actually t(n-1) where t is the thickness and n is the refractive index of the filter) from the spacing between corrector and ccd. You will also need to refocus. Regards Andrew

Not wishing to complicate things but I have been pondering what one should do if you have a corrector that is required to be a certain distance from the detector and you place a filter between them. If you assume that corrector is designed to have a certain optical path distance ( i.e. geometric distance * refractive index) from the corrector to CCD then if you place a filter of say Xmm thickness and of refractive index n between them, without increasing the geometric distance, it adds X*(n-1) to the optical path length - assuming the index of air to be ~1. So the geometric distance between the corrector and CCD needs to be reduced by this amount does it not so the total optical path length stays at the design distance! Refocusing will also be required. I still stand by what I posted before but for this particular point. Regards Andrew

I knew I had it somewhere - attached from Telescope Opitcs Rutten & van Venrooij - Andrew

I am never clear what adding or subtracting backfocus means. I still believe the glass plate results in the focus being further from the objectve. I attach a sketch that show the effect. While not accurately to scale the key point is the light is refracted towards the normal as the light enters the plate and away when it leaves. This means the light converges less quickly when in the plate. Regards Andrew

If the filter is between the flatner and the chip I would increase the flatner to chip distance to account for the filter as specified above in a previous post.Regards Andrew

The filter acts to push the focus further away from the objective/mirror so to reach focus you need to move the CCD outwards Regards Andrew