# Functioning of a compound microscope

"The diameter of the eyepiece is greater than that of the objective. This helps to collect more light and gives a brighter image"

I thought that increasing the aperture of a lens doesn't increase the brightness of a virtual image. Yes, it will increase the brightness of a real image caught on a screen, but not a virtual image (I think) So how will increasing the diameter make the virtual image brighter?

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Formation of image when object is at infinity

I know how a beam of light rays parallel to the principal axis, after refraction through a convex lens, passes through the focus. But what about formation of an image. In my book the light rays are not parallel to the principal axis (they are parallel to each other though) so the are incident obliquely on the mirror and they say the image is formed in the focal plane. But what about the height of the image? How much smaller is it? I though it should be point sized? And how do you know which light ray is coming from which point on the object since all the rays are going to be parallel?

Drakkith
Staff Emeritus
Every point on the object emits light rays that are made parallel to each other from the one point they are emitted from. A point that is 0.1 mm away from another point will not have light rays parallel to the other points light rays.

Also, increasing the diameter of the eyepiece should make the image brighter in all cases as long as the focal length does not change. (As long as the light cone is bigger than the eyepiece) Why do you think the virtual image shouldn't be brighter?

sophiecentaur
Gold Member
If the eyepiece is wider than your pupil, how can light from the edges get into your pupil and make the retinal image brighter? A wider objective lens (aperture) will gather more light and, given the right layout, this can make the real, intermediate, image brighter but your eye pupil represents a limiting aperture in that part of the optics.

Drakkith
Staff Emeritus
If the eyepiece is wider than your pupil, how can light from the edges get into your pupil and make the retinal image brighter? A wider objective lens (aperture) will gather more light and, given the right layout, this can make the real, intermediate, image brighter but your eye pupil represents a limiting aperture in that part of the optics.
Ahh, that must be why the "virtual" image doesn't get brighter! Remove your eye and adjust the focus so that the light is focused onto a piece of paper and you will get a brighter image with a larger aperture. That sound right?

I don't have a telescope to be playing around with so I can't be trying all this out :/ But as for the images getting brighter, here's what I think:-

You're using a convex lens to form a real image which is captured on a screen. I get how the image on the screen will be brighter if you increase that aperture. But if you were not capturing it on a screen I don't think you could call it brighter. With an increase in aperture, each point on the real image will emit light in more directions, but the "intensity" remains the same. It just emits light in more directions. And it's obvious how if it emits light in more directions, then if you were to capture the image on a screen it would be brighter.

For a virtual image it's the same thing. With an increase in aperture, same amount of light in all directions, but now in more directions than before. And you can't capture a virtual image on a screen, so you can't get a brighter image.

sophiecentaur
Gold Member
I don't have a telescope to be playing around with so I can't be trying all this out :/ But as for the images getting brighter, here's what I think:-

You're using a convex lens to form a real image which is captured on a screen. I get how the image on the screen will be brighter if you increase that aperture. But if you were not capturing it on a screen I don't think you could call it brighter. With an increase in aperture, each point on the real image will emit light in more directions, but the "intensity" remains the same. It just emits light in more directions. And it's obvious how if it emits light in more directions, then if you were to capture the image on a screen it would be brighter.

For a virtual image it's the same thing. With an increase in aperture, same amount of light in all directions, but now in more directions than before. And you can't capture a virtual image on a screen, so you can't get a brighter image.
You need to think carefully about the relevant factors here and there are two situations to consider. You are right that a bigger objective lens (simple convex) will put more light onto a real image in all circumstances. This will show up as an on-screen brightness pro-rata with lens area. You can look at this image from behind and, for small objective diameters, all the light cone from the objective will get into your pupil so the brightness will increase with objective diameter up to that point. Beyond that, light will just spill out, around your pupil - again, as you say. For parts of the image that are far from from the principle axis, still less light will get into your pupil so you will get 'vignetting' or darkening of the outside of the image.

A suitable eyepiece (basically another convex lens), as well as placing the image at a more convenient distance away, will intercept all / more of the cone of light from the objective lens (over a range of angles) and alter the cone angle so that it all gets into your pupil.
A properly designed telescope will ensure that the objective is no wider than the eyepiece can cope with. This means that the focal lengths need to be appropriate or you will be wasting objective area and getting vignetting. Astronomical telescopes are often made with long barrels and this must help to avoid light loss for a particular objective diameter.

Vignetting is present in many (cheap) camera (multi-element) lenses which have a large front optic but inadequate internal optics to make sure that light gets to all parts of the film when the maximum aperture is used (when the cone from the objective is too wide). The way the manufacturers specify naff lenses is criminal!

Andy Resnick
There may be some confusion here- the physical diameter of a lens does not always equal the size of the entrance pupil, aperture, or exit pupil. In terms of eyepiece design, what matters is that the exit pupil is located at the entrance pupil of your eye, which is close to (but not equal to) your iris. If the eyepiece exit pupil is too large or too small, as sophiecentaur mentions, light is wasted or vignetting occurs.

Now- matching the eyepiece to the objective. Lens throughout is specified by the product of the entrance pupil diameter and the numerical aperture- the etendue. The etendue is a constant unless vignetting occurs. Because the exit pupil of the objective lens is much larger than the entrance pupil, the exiting cone of light is much smaller than the numerical aperture specification, and the numerical aperture of the eyepiece does not have to be large. In any case, the eyepiece typically specifies both a magnification and an object size (field of view). For example, I generally use a 10X/18mm eyepiece, but I also have a 15x/12mm eyepiece. You can see that the etendue of both the 10/18 and 15/12 are identical, so as long as the objective lens etendue is large enough, the image will have the same brightness- even though the 'diameter' of the 10x eyepiece is much larger and the fields of view are different.

sophiecentaur
Gold Member
@Andy Resnick
Good information there, thanks.
I seem to remember reading that a galilean telescope, which uses a concave eyepiece lens has very poor 'light gathering' and is limited just by the pupil of the eye, whatever the aperture of the objective. How would you specify such an eyepiece in the terms of that last post? What would its etendue be?

Andy Resnick
<snip>How would you specify such an eyepiece in the terms of that last post? What would its etendue be?
Tough to say, without an optical specification. For example, let's make a Galilean telescope out of a 100mm diameter front element, operating at f/10, and the rear eyepiece is a concave lens, 50mm focal length. The (angular) magnification is 1000/50 = 20X, given by the ratio of focal lengths. The numerical aperture of the objective lens is 0.05, and so the etendue of the objective lens is 3.14(50^2)*(0.05)^2 = 19.6. I didn't specify the diameter of the eyepiece, but the illuminated area of the eyepiece can be estimated by geometry to be a circle of diameter 100*(50/1000) = 5mm, so the eyepiece is also operating at f/10. However, the etendue of the eyepiece is only 0.05- much less than the objective lens. What is happening is that the eyepiece restricts the field of view of the objective lens- this is the major deficit of the Galilean eyepiece.

My optical design books are away in my office- I was hoping to be more explicit about field of view calculations to compare Galilean and Keplerian eyepieces, but that will have to wait....

Drakkith
Staff Emeritus
What the heck is etendue? I looked it up and I still have no idea what it is.

sophiecentaur
Gold Member
What the heck is etendue? I looked it up and I still have no idea what it is.
Doesn't wiki tell you enough? You don't need to follow the Maths to grasp what it's about. It seems, to me, to be a very helpful concept.

Drakkith
Staff Emeritus
Doesn't wiki tell you enough? You don't need to follow the Maths to grasp what it's about. It seems, to me, to be a very helpful concept.
No, I have absolutely no idea what wiki is saying.

sophiecentaur
Gold Member
No, I have absolutely no idea what wiki is saying.
That article starts off by defining what etendue is and it's easy enough to calculate. You multiply two understandable (?) quantities together. That could be enough but then they show you examples of how etendue applies. The fact that it never reduces is useful to know.
Is it the Maths that gives the problem? Some things just can't be done without Maths.

Andy Resnick
Bleah- I spent a few hours on this today- with a headache for my reward. Surprisingly, these simple two-lens systems are quite difficult to analyze because they are afocal systems- the usual ray tracing method to locate pupils and stops fails completely, and *none* of my optical design books have anything to say on the subject. They all discuss telephoto lenses, which are focal versions of Galilean telescopes and Kingslake mentions that camera viewfinders are often reversed Galilean telescopes... thanks, dude. To summarize:

I worked with both a Galilean and Keplerian telescope, both with an objective lens f1 = 1000 mm, diameter = 100mm, and an eyepiece that was either f2 = -50mm (Galilean) or 50mm (Keplerian). The lenses are spaced by the sum of their focal lengths, 950mm (G) and 1050mm (K). They have the same magnification (20), but G has an upright image (M = 20) while K has an inverted image (M = -20). For both, the magnification is mostly on the objective- M = 400 for f1, while M = +/- 0.05 for the eyepiece- that's surprising, I would have expected the optical power to be more equally distributed. Perhaps it was a sign of the problems to come....

The etendue (E) is related to the Lagrange invariant (H) for an optical system, and both are measures of the throughput. The Lagrange invariant can be derived from Gaussian optics, relating the magnification to aperture angle ratio. In afocal systems, since the aperture angles are zero (the objects are at infinity), H is written as H= nyq, where 'n' is the refractive index, 'y' the marginal ray height at the entrance pupil, and 'q' the field angle. If the entrance pupil is the objective lens, then both K and G have H = 50*q. E and H are equivalent- E = pi^2 H^2 for circular systems.

So calculating either H or E appears trivial- all I need to do is calculate the field angle q for each telescope. That's where I failed completely. I was able to calculate the marginal ray height at the eyepiece (2.5mm for both), but could not locate the aperture stop for either K or G; the aperture stop is used to calculate the entrance pupil (I was getting entrance pupils 20m in front of the objective when it should have been located *at* the objective) and exit pupil (should be located near the eyepiece), and that is used to calculate the field of view.

Web searches didn't help either- all I found was that a Galilean telescope has field angles of about 10 arcminutes (I calculated 9 arcmin for G), and that Keplerian telescope has 'a much larger field of view', although my calculations provided a field of view of 8 arcmin for K. In any case, comparing the etendue/Lagrange invariant for K and G shows that K will have a larger value because the field of view is larger.

Off to the pub, maybe that will help....

sophiecentaur
Gold Member
You did very well there and thanks for the effort. This isn't my field but I got the bit about the 'afocal' problem. Obviously, all the work (and money) is in camera lenses so the analysis has presumably gone that way.
I'm surprised at such a narrow field for a Galilean telescope - but perhaps you were referring to 'serious ones'. My Dad made one (more like an opera glass) about 20cm long and that had a wider field than less that a degree.
That business of the aperture of the Galilean telescope aperture being the same as the eye rings a bell too.
You can have a rest now.

Andy Resnick
...Several pints of "Christmas Crack" (google it) later and I figured it out. Here goes:

Two telescopes (Keplerian, K, and Galilean, G) are constructed using a 1000mm focal length objective and a 50mm negative (G) or positive (K) lens. The lenses are spaced at f1 + f2 = 950 (G) or 1050mm (K) and the magnification is 20 (G) or -20 (K). The key was to set the entrance pupil at the objective lens.

The entrance pupil is the hole through which all light must enter the optical system, and the exit pupil is the hole through which all light must leave the system. The entrance pupil diameter is 100mm as before, and the exit pupil diameter is 5mm, as per the magnification. *Where* is the exit pupil?

The using the thin lens equation on the eyepiece with the objective lens as the object means the image plane is the exit pupil! That is located at -47.5mm (G) or 52.5mm (K): the exit pupil in a Galilean telescope is *inside* the telescope, while the exit pupil for the Keplerian telescope is located outside of the telescope.

For K, there is no additional complications- place the entrance pupil of your eye 52.5mm away from the eyepiece, and you will see an image. The K exit pupil is well-matched to the size of your iris, and since the field of view of your eye can go up to 30 degrees or more, the field of view of the telescope can (in principle) be as high as 1.5 degrees. In practice, the field of view is less to provide a well-corrected field of view, but 15 degrees is very reasonable, leading to an object field of view of 0.75 degrees, so the Lagrange invariant for this is 37.5.

The situation is drastically different for G- now, you can't access the exit pupil- the best you can do is mash your eye up against the lens and even then, there will be severe vignetting. If we keep your eye position for G the same as K, it means the angle of the chief ray (which defines the field of view) leaving the center of the exit pupil must be small enough to pass through the edge of the entrance pupil of your eye (2.5mm field height)- this is 2.5 degrees, which corresponds to 7 arcmin object field of view, with a Lagrange invariant of 5.8, considerably smaller than K.

Surprising that changing a single element can have such a drastic effect!

Now I'm going to pass out....