Focal Adjustment to Compensate for Refractive Index

Have you drawn a ray diagram?

Here's one to save you the effort:

(Ray paths through the lens are approximate - and don't matter anyway, because it's the effect of the flat glass that we're interested in).

Look out for the extra chromatic aberration too.

You seem to be agreeing with what Usbport is saying. The lens focal point (your down arrow) has moved left so I have to move the lens right to compensate.

Correct.

That's what we've been doing but a suggestion has been made that it was wrong.

We did a practical check using our vision measuring system. We took a 5.6mm thick quartz plano optic & measured the thickness by focusing on an ink mark on the top & bottom faces. The measured thickness was 3.8mm which makes sense if you multiply the 3.8 by the quartz refractive index of 1.74.

This seems to me to confirm what has been said here.

Yes - it ties up. Here's the ray diagram for this case:

I haven't drawn a diagram but thanks for yours.

Light slows down in any medium with n>1; thus the convergence of the rays to a focal point is delayed by the index of the material and its thickness. The higher the value of n and the greater the thickness, the longer the beam will take to reach focus.

So in general, placing a filter in a converging light path before it has reached the focal point will cause the back focal distance to slightly increase.

"will cause the back focal distance to slightly increase"

So I have to move the lens away from the sensor to compensate?

Yes. Looks like you've gotten lots of help.

I think I'm convinced by all of this but we're going to set up a practical test to prove it to my workmates. I'll report back on the result.

Be interesting if you come up with a result different from this axiom in part 1 of Newton's Opticks (2^nd edn., 1718):

Defin. IV.

"The Angle of Incidence is that Angle, which the Line described by the incident Ray contains with the Perpendicular to the reflecting or refracting Surface at the Point of Incidence."

Defin. V.

"The Angle of Reflexion or refraction, is the Angle which the line described by the reflected or refracted Ray containeth with the Perpendicular to the reflecting or refracting Surface at the Point of Incidence."

Axiom IV.

"Refraction out of the rarer Medium into the denser, is made towards the Perpendicular; that is, so that the Angle of Refraction be less than the Angle of Incidence."

If our result contradicts Newton I think I'll be asking them to repeat the test.

Your reasoning is correct.

There is a perfect "real world" example I discovered about CCD SLR cameras. When we put the lenses from our old film SLR on the newer digital (CCD) cameras, all of the photos were the slightest bit fuzzy. We couldn't figure it out (on our own) for the longest time. My wife was quite frusrated with her new (expensive!) camera.

A little searching on the 'net revealed that the focal length was changed by the protective cover ("glass") over the CCD, and different "optimized" lenses would need to be used.

One would think that the industry would have made the change in the camera so all of the old lenses could continue to be used. Oh, really? Guess you'll have to buy all new lenses...hmmm...our bad.

The CCD cover glasses are typically about 0.5-0.7mm thick glass so the back focal length adjustment would be about 0.2mm, you might get away with using a spacer between the lens & lens mount.

She was worth the new lenses. We needed higher quality ones anyway when she got a new camera (Canon 40d). I have the old one now.

Our test set-up confirmed what has been suggested in the posts here. It seems that the way we have adjusted focal distances on our cameras has been correct but has been done without really understanding the rationale behind it. Now, if I'm asked to justify what I'm doing I can explain it by reference to the optic material thickness & the refractive index.

Thanks for all the comments.

You've got it backwards. Placing a glass window will lengthen the optical path, not shorten it. If the glass is 1 mm thick and has a refractive index of 1.5, it will add:

1 mm - (1 mm / 1.5) = .333 mm

to the optical path length.

Right...as in post #3 (visual aid).

You are replying to the OP (original post), in which the question posed was open (i.e. no assertion was made as to whether the optical path would be longer or shorter). The OP (original poster) could not have "got it backwards".

If you were directing your reply to another post, please indicate which.