Finding dust in the path of the image formation

[Ecki]

Dear friends,

for a little while I was having problems with my 63x objective - there were shadows in the image. My observations were as follows:

1) Images with the 40x/1.30 are without problems
2) Images with the 63x/1.40 have shadows
3) All lenses and DIC prisms are clean

To identify the lens that gives problems you need to turn lenses/optical components. When you turn the right lens, the dust turns as well. But to follow this procedure I needed to see the dust better than just some shadows!

Here is a little trick that I got from a friendly Zeiss technician that I hope you will find helpful as well.



You can see the shadows left of the middle. Now I lowered the condensor to reduce the NA.





Lowering the condensor makes the dust appear!

Now it was fairly easy to identify the lens that has dust - it was the camera adapter lens. 5 minutes later the shadows were passé!



I was not aware that reducing the NA of the condensor will enhance the contrast to show dust in the path of the image formation.

Why did the 63/1.40 show the shadows and not the 40x/1.30? This is because the condensor has an NA of 0.90 and the difference between the objective NA and the condenser NA is big. The same effect made the dust clearly visible when I lowered the condensor - increasing the difference between lens NA and condensor NA!

I hope you will find this trick helpful - just in case

Regards
Ecki

[rjlittlefield]

Ecki,

Thank you for passing along this trick.

I was not aware that reducing the NA of the condensor will enhance the contrast to show dust in the path of the image formation.

I think it is a matter of more DOF. Notice that when the condenser is dropped and the dust is more visible, then your purple and brown subjects have a lot more DOF also. Anything that stops down the system will make OOF dust more apparent.

Why did the 63/1.40 show the shadows and not the 40x/1.30? This is because the condensor has an NA of 0.90 and the difference between the objective NA and the condenser NA is big. The same effect made the dust clearly visible when I lowered the condensor - increasing the difference between lens NA and condensor NA!

It is an interesting observation that the 63/1.40 shows the shadows and not the 40x/1.30. I think this is telling us something more than just the difference between lens NA and condensor NA.

At least with the model in my head, an NA 0.90 condensor that is otherwise perfectly matched to the objectives will simply stop both objectives down to NA 0.90 and the OOF dust will be equally blurred in both cases (ignoring difference in magnification).

However, a condensor that is not perfectly set up may not deliver its full NA because steeply angled rays of light fail to enter the objective. This is what happens when you drop the condensor. What could be happening is that your 63X is not as well matched to the condensor as your 40X, so that the 63X never gets up to even the nominal NA 0.90.

You might explore this by looking at exposure times with the two objectives. If both the 40X and 63X are running at NA 0.90, then difference in exposure times should be due only to the difference in magnifications, so a ratio of (63/40)^2 = 2.48. Any larger ratio than that must be due to either NA or absorbance in the glass.

--Rik

[Ecki]

Rik,

the distance d of two points that can be separated with a microscope is

d = 1.22 x wavelength / 2 x A

where A is the NA. In the case of the lens and the condensor having a different NA, NA = (NA condensor + NA lens)/2. Therefore we can use

d = 1,22 x wavelength / (NA condensor + NA lens).

Taking 550 nm for visible light that gives us a theoretical d = 240 nm for a system with NA lens 1.40 and NA condenser 1.40 and d = 290 nm for a system with NA lens 1.40 and NA condenser 0.90 - just 50 nm difference for d.

This is why there is no 1.40 condensor available for the Axioscope A.1. The small condensor aperture has the same effect as an aperture iris - improving contrast and DOF but of course reducing resolution d.

I think it is a matter of more DOF. Notice that when the condenser is dropped and the dust is more visible, then your purple and brown subjects have a lot more DOF also.

As the dust was 30 cm above the object carrier I think this is another effect and not DOF.

Regards
Ecki

[rjlittlefield]

As the dust was 30 cm above the object carrier I think this is another effect and not DOF.

Let me try again with different words.

The first point behind my post is that the appearances of all OOF obstructions in the optical path are determined by the NA of the optical system. Reducing the NA makes light cones narrower throughout the system. This reduces the blur radius for anything that is out of focus, no matter where it is located. For things near the subject plane this gives more Depth of Field; for things near the sensor it gives more Depth of Focus; and for things elsewhere in the system it has the same effect even if we don't know what to call it. I used "DOF" as a shorthand for the whole lot.

My second point was that the appearance of OOF dust should be the same with both objectives, if they are both accepting light at NA 0.90. That point was actually not correct.

What I overlooked is that in back of the objective, the system NA gets scaled by the magnification. As a result, behind an objective at 63X and NA 0.90 the system is NA 0.0143, while behind 40X and NA 0.90 the system is NA 0.0225.

The difference between 0.0143 and 0.0225, caused by magnification, is enough to explain why the dust would be a problem in one case but not the other.

The relationship between lens NA and condensor NA does not play into any of this directly, but of course there is a definite correlation because higher lens NA and higher magnification usually go together.

Thank you for the further discussion.

Best regards,
--Rik

[Pau]

Ecki and Rik,
Very useful info and intersting discussion.

[Ecki]

Rik,

of course I had to defend my combination of 0.90 condensor and 1.40 lens But I think it is pretty interesting that the actual loss of resolution is pretty small - 50 nm.

The first point behind my post is that the appearances of all OOF obstructions in the optical path are determined by the NA of the optical system. Reducing the NA makes light cones narrower throughout the system. This reduces the blur radius for anything that is out of focus, no matter where it is located. For things near the subject plane this gives more Depth of Field; for things near the sensor it gives more Depth of Focus; and for things elsewhere in the system it has the same effect even if we don't know what to call it. I used "DOF" as a shorthand for the whole lot.

I think this is the correct description. It also explains why dust close to the sensor creates the biggest disturbances.

What I overlooked is that in back of the objective, the system NA gets scaled by the magnification. As a result, behind an objective at 63X and NA 0.90 the system is NA 0.0143, while behind 40X and NA 0.90 the system is NA 0.0225.

This I don't understand. How do you calculate those numbers?

Regards
Ecki

[rjlittlefield]

But I think it is pretty interesting that the actual loss of resolution is pretty small - 50 nm.

True, but on the other hand... The calculation can also be done as a percentage. Using the (NA condensor + NA lens) formula, a 1.40 lens and a 0.90 condensor loses 17.8% compared to 1.40 and 1.40. Going to a 0.90 lens would lose only 21.7% more, and would allow to work dry. It is always a matter of tradeoffs.

But really my concern was that there might be some alignment mismatch between the condensor and the 63X lens that would cause more of the edge rays to not enter the lens. After more consideration, I think this is not likely.

How do you calculate those numbers?

It's very simple: exit NA is always equal to entrance NA divided by magnification, so 0.0143 = 0.90/63 and 0.0225 = 0.90/40.

This is a very handy formula for estimating diffraction effects, especially for photographers who are used to thinking in terms of "effective f-number" as seen by the camera. In that case we note that effective f-number = 1/(2*NA) and combine the two formulas to get that

effective f-number at camera = magnification/(2*entrance_NA).

For example at 10X and NA 0.25, the system gives effective f/20 (=10/(2*0.25)) as measured at the camera. This is already into diffraction territory with a 15 megapixel APS-C sensor. But 63X NA 1.40 is only a little bit worse, effective f/22.5, so this explains why your 63X objective still gives such sharp pictures.

--Rik