# Problem understanding magnification

So there are two facts that I know:

1) The farther the image is formed from a mirror/lens, the bigger it is.
2) The more powerful a lens is, the closer the image is formed assuming the object at a fixed distance.

I can't reconcile these two facts. Common sense says that the stronger the lens that is used, the greater should be the magnification. Yet the size of the image depends on its distance from the lens, so if I'm going to use a thicker lens, it will form an image nearer to it than a thinner lens would. Nearer means smaller, so actually, we should use as thin a lens as possible to achieve max magnification.

DEFINITELY not one of my better ideas...

Related Classical Physics News on Phys.org
jambaugh
Gold Member
You need to consider the difference between a virtual and actual image.
When you burn ants with a magnifying glass you bring the sun into focus on a small point. The more powerful the lens the more tightly focused the light... thus smaller the actual image of the sun relative to its true size. You can in fact express the magnification in terms of the ratios of the distances of object and image to the lens.

But in using a magnifying lens (the very same one with which you may have burnt ants) to magnify the ants you are looking through the lens. You have in effect a compound lens in combining your eye's lens with the magnifying glass. You see with your eye a virtual image of the ant which is larger than the ant itself.

To appreciate virtual images you need to consider the reversal symmetry of object and image. Take the case of focusing the sun to a small circle on the back of an ant. If you remove the sun and place a screen that far off, and if you replace the sun's image with an incandescent dot, you will see the object and image swap, the small dot produces a huge image very far away. In general if Object A a distance $a$ away from a lens produces Image B at a distance $b$ from the lens then an object B would produce an Image A if you reverse the direction light is traveling through the lens.

With that in mind consider your eyeball looking at an ant as if an image of the ant on your retina becomes the object projected by your eye out on the ground. Run that projection through a (concave) lens (which is farther than its focal length from your eye) and it will produce a smaller image ant at a shorter distance away. So to compensate you start with a bigger ant picture on your retina and change the focus of your eye's lens so that this bigger ant projects out of your eye to exactly the original ant.

Now we reverse the reversal and you see that the actual ant, as seen through first the lens in your hand and then your eye lens must form a bigger image on your retina, it i.e. looks bigger... (and farther away in terms of your eye's lens focusing on it) than the original. This looking through the lens image is called a virtual image.

Hope my verbose explanation was helpful.

ViolentCorpse
You need to consider the difference between a virtual and actual image.
When you burn ants with a magnifying glass you bring the sun into focus on a small point. The more powerful the lens the more tightly focused the light... thus smaller the actual image of the sun relative to its true size. You can in fact express the magnification in terms of the ratios of the distances of object and image to the lens.

But in using a magnifying lens (the very same one with which you may have burnt ants) to magnify the ants you are looking through the lens. You have in effect a compound lens in combining your eye's lens with the magnifying glass. You see with your eye a virtual image of the ant which is larger than the ant itself.

To appreciate virtual images you need to consider the reversal symmetry of object and image. Take the case of focusing the sun to a small circle on the back of an ant. If you remove the sun and place a screen that far off, and if you replace the sun's image with an incandescent dot, you will see the object and image swap, the small dot produces a huge image very far away. In general if Object A a distance $a$ away from a lens produces Image B at a distance $b$ from the lens then an object B would produce an Image A if you reverse the direction light is traveling through the lens.

With that in mind consider your eyeball looking at an ant as if an image of the ant on your retina becomes the object projected by your eye out on the ground. Run that projection through a (concave) lens (which is farther than its focal length from your eye) and it will produce a smaller image ant at a shorter distance away. So to compensate you start with a bigger ant picture on your retina and change the focus of your eye's lens so that this bigger ant projects out of your eye to exactly the original ant.

Now we reverse the reversal and you see that the actual ant, as seen through first the lens in your hand and then your eye lens must form a bigger image on your retina, it i.e. looks bigger... (and farther away in terms of your eye's lens focusing on it) than the original. This looking through the lens image is called a virtual image.

Hope my verbose explanation was helpful.
Thanks for so detailed an explanation. Very interesting to learn about the symmetry between object and image. But I'm still confused about the relation between the power/focal length of a lens and magnification. E.g a simple telescope or a compound microscope first forms a real image, and then a second lens forms an enlarged virtual image of that real image. Now, if a thicker objective lens is used, it will form a real image nearer to it and thus smaller. The virtual image formed by the eyepiece then in this case, will also be smaller because its object (the real image) is now smaller.

In short, I'm unable to escape the conclusion that shortening the focal length or increasing the power will result in smaller magnification, but this sounds absurd.

Anyone?

Drakkith
Staff Emeritus
You have a slight misunderstanding of how magnification is calculated. From wiki: Optical magnification is the ratio between the apparent size of an object (or its size in an image) and its true size, and thus it is a dimensionless number.

Consider the angular distance (apparent size) between two objects in the image field, A and B, with no optical system present. If we insert an optical system, the magnification is determined by the angular distance between A and B at the focal plane. In the image below, red rays are from A and yellow rays are from B. As the lens power is increased, the focal length is reduced and rays from both A and B are brought to focus closer to the lens. This means that the distance between A and B at the focal plane is also reduced and so is magnification. So the longer the focal length of the lens, the larger its magnification in this sense.

Note that the above picture is for when the distance between each object and the objective lens is much larger than the focal length of the objective. In the image below we have a case where the distance between the object and the lens is smaller than the focal length. In such a case, the light will not form a real image, but a virtual one. When viewed by the eye, the apparent diameter of the object is larger than normal, as shown. Now, if we increase the power of the lens, the rays entering the eye would be converging, and we would have to decrease the power of the eye in order to still bring the rays to focus at the retina. This results in a drop in magnification since magnification in a two-element system is the focal length of the objective divided by the focal length of the 2nd element. But, in this simple case, we can simply move the lens closer to the object and leave the power of the eye unchanged. Since we are moving the lens and eye closer to the object, the apparent size of the object itself increases, which is then magnified by the optical system to form a larger image. In other words, the image gets larger because we can move closer to the object and still maintain focus.

E.g a simple telescope or a compound microscope first forms a real image, and then a second lens forms an enlarged virtual image of that real image. Now, if a thicker objective lens is used, it will form a real image nearer to it and thus smaller. The virtual image formed by the eyepiece then in this case, will also be smaller because its object (the real image) is now smaller.
Not necessarily true. Two different objectives can be of different thicknesses yet have the same magnification/focal length. For a simple thin lens, it's not the distance from the surface of the lens, but from the center of the lens that matters. For a thick lens, the focal length depends on multiple factors, some of which you can read about here: https://en.wikipedia.org/wiki/Focal_length#General_optical_systems

Note that it's much better to think about magnification in terms of focal length and angular distances rather than linear distance from a lens. This is especially true in multi-element and/or reflecting systems.

ViolentCorpse
Thank you, Drakkith for being so descriptive. I really appreciate it. I understand it better now.

I'm still having difficulty with compound system of lenses, though. I'm afraid I couldn't express my question very well.

Not necessarily true. Two different objectives can be of different thicknesses yet have the same magnification/focal length. For a simple thin lens, it's not the distance from the surface of the lens, but from the center of the lens that matters. For a thick lens, the focal length depends on multiple factors, some of which you can read about here: https://en.wikipedia.org/wiki/Focal_length#General_optical_systems
When I said "thicker" I was trying to say "a lens with shorter focal length/more power". It's the relation between image distance and magnification that I'm concerned about.

In the case of a simple magnifier, there is just a virtual image as you explained. But in a compound microscope/telescope, there is a real image involved as well. Now, if an objective lens of higher power/shorter focal length is used, would the real image formed be nearer and smaller than before (when the lens was of a lesser power)? That's my real question.

I feel bad asking for more after you've already obliged me with such a lucid, detailed response. Alas, I'm a bit of a dimwit.

Thank you so much, once again!

Drakkith
Staff Emeritus
In the case of a simple magnifier, there is just a virtual image as you explained. But in a compound microscope/telescope, there is a real image involved as well. Now, if an objective lens of higher power/shorter focal length is used, would the real image formed be nearer and smaller than before (when the lens was of a lesser power)? That's my real question.
Yes, if you increase the power of the objective lens, then the focal length is decreased and the real image is formed closer to the lens and is smaller. Note that the overall magnification may or may not increase in a compound optical system. For example, playing around with the microscope magnification page here, I find that decreasing the focal length of the objective lens (increasing its power) increases the overall magnification. But for a telescope, decreasing the focal length of the objective decreases overall magnification. I'm honestly no quite sure why the difference between the two exists.

ViolentCorpse
jambaugh
Gold Member
Something to note, the magnification is not a fixed function of focal length. You can, with the same lens form an image of an object which is bigger, smaller or equal sized.
ViolentCorpse,
it seems to me from your comments that you are implicitly assuming this is not the case and that magnification is a fixed function of focal length.

This is only the case when you fix one of the other variables as well, either the distance from object to lens (typical of telescopes) or the distance form image to lens (typical or a microscope) or thirdly, fixing the object to image distance (typical of simple magnifier use).

Note that for a telescope, your primary image is smaller than the secondary (magnification is the ratio of the corresponding distances to the lens). Weakening the primary in the sense of using a longer focal length will in fact increase magnification in that it will decrease the de-magnification of the primary image. (You then use the secondary to form a virtual image larger than the actual object which your eye then resolves as a larger retinal image). You then use a secondary closer to the primary than the focus but reverse mapping to virtual image we see from the image of the primary. Assuming the secondary virtual image is at or near infinity the primary image is at or near the focal length of the secondary (but past the secondary). Given a fixed distance to eye, the secondary magnification will be larger for shorter focal lengths.

(Side note: you might imagine varying the distance from secondary to eye but that would severely limit field of view. You get some of that in rifle scopes since you generally don't want a rifle kicking the scope into your eye socket, there must be some distance there.)

Below I work out (without checking my work mind you) the formula for simple telescope magnification...

Ok now considering the telescopes total magnification in the form of say a virtual image at the same axial position as the original object give or take a small amount on the order of the telescope lengths... let $R$ be the distance to object, $r_1$ be the distance to primary image, $r_2$ be distance from secondary lens to primary image. Let $f_1,f_2$ be the primary and secondary focal lengths.

$$\frac{1}{R}+\frac{1}{r_1}=\frac{1}{f_1}$$
so
$$r_1 = \frac{1}{\frac{1}{f_1}-\frac{1}{R} } = \frac{R\cdot f_1}{R-f_1 } = f_1\cdot \frac{1}{1-f_1/R}$$
The (de)magnification is $r_1/R = \frac{f_1}{R-f_1} \approx \frac{f_1}{R}$ since $R >> f_1$.
Then considering the secondary lens we essentially use it to project the primary image to a secondary image located back at the original object (minus the telescope length)...
so as before...
$$\frac{1}{r_2}+\frac{1}{R}=\frac{1}{f_2}$$

$$r_2 = \frac{R f_2}{R - f_2}$$
note however that the secondary magnification is:
$$\frac{R}{r_2} =R \frac{R - f_2}{R\cdot f_2} = \frac{R}{f_2} - 1 \approx \frac{R}{f_2}$$

Total magnification is then the product of these which comes out to about the ratio of the focal lengths: $Mag\approx \frac{f_1}{f_2}$. Stronger eyepieces increase magnification as your intuition might assume, but note also weaker primary lenses also increase magnification.

[I may have messed up not accounting for virtual images having negative distances but that would be a sign issue with a term we neglect when assuming the R>>f condition.]

ViolentCorpse and Drakkith
Something to note, the magnification is not a fixed function of focal length. You can, with the same lens form an image of an object which is bigger, smaller or equal sized.
ViolentCorpse,
it seems to me from your comments that you are implicitly assuming this is not the case and that magnification is a fixed function of focal length.

This is only the case when you fix one of the other variables as well, either the distance from object to lens (typical of telescopes) or the distance form image to lens (typical or a microscope) or thirdly, fixing the object to image distance (typical of simple magnifier use).
Yes, actually, I was assuming that the object distance to lens was fixed.

$Mag\approx \frac{f_1}{f_2}$. Stronger eyepieces increase magnification as your intuition might assume, but note also weaker primary lenses also increase magnification.]
Doesn't this formula say the opposite? Stronger primaries and weaker eyepieces to increase magnification?

At any rate, you and Drakkith have been most helpful. Thank you so much guys! ^_^

Drakkith
Staff Emeritus
Doesn't this formula say the opposite? Stronger primaries and weaker eyepieces to increase magnification?
Nope. As the primary increases in power, the focal length decreases, so the ratio gets smaller. Conversely, as the focal length of the eyepiece gets shorter, the ratio increases.

Example:

Focal length of primary: 1000 mm
Focal length of eyepiece: 20 mm
Magnification: 1000/20 = 50x

Changing the focal length of the primary to 800 mm: 800/20 = 40x
Changing the focal length of the secondary to 10 mm: 800/10 = 80x

ViolentCorpse
Nope. As the primary increases in power, the focal length decreases, so the ratio gets smaller. Conversely, as the focal length of the eyepiece gets shorter, the ratio increases.

Example:

Focal length of primary: 1000 mm
Focal length of eyepiece: 20 mm
Magnification: 1000/20 = 50x

Changing the focal length of the primary to 800 mm: 800/20 = 40x
Changing the focal length of the secondary to 10 mm: 800/10 = 80x
Oops. I mistook focal length for power.

Thanks again, Drakkith!

There are a few unrelated questions I want to ask. Answer them only if it in no way imposes on you, please.

1. Do spectacles form virtual images? Or do they just help focus a real image on the retina?
2. An object placed at the focal length of a convex lens produces no image at all. Yet, there is no point in front of our eye where an object would suddenly disappear from our sight. What's going on?

Thanks a ton, guys!

Drakkith
Staff Emeritus
1. Do spectacles form virtual images? Or do they just help focus a real image on the retina?
That depends. Spectacles are corrective lenses, meaning that they are designed to correct for aberrations introduced by imperfections in the eye. Sometimes they are slightly positive, sometimes negative, and sometimes they have little to no optical power, but correct for aberrations other than nearsightedness and farsightedness. A lens that corrects for astigmatism has some optical power, but only in one axis.

2. An object placed at the focal length of a convex lens produces no image at all. Yet, there is no point in front of our eye where an object would suddenly disappear from our sight. What's going on?
Sure there is. An object place around 15-25 mm from the eye will be at one focal length and the light will not form an image on the retina. When we say that an image isn't formed, we mean that the light leaving a lens/mirror is parallel to it itself instead of converging or diverging. However, you can still see an 'image' on your retina or on a camera lens, it's just extremely blurry.

I can't find a proper image, so we'll have to improvise. In the image below, imagine that the light is leaving an object at the focal point instead of converging on it. The light rays are refracted so that they emerge parallel to themselves. Now, imagine your retina is on the left. All of the light will fall on your retina within the diameter of that column of light, and none will be outside of it. Light emitted from a point just below the illustrated focal point will form its own column of parallel light which will mostly overlap the previous column except for a small sliver on the edge. So if you put your finger very close to your eye you will see a large, blurry blotch where most of the different columns of light from your finger are overlapping, but the edge of your finger will be an extended blur which you can see more distant objects through.

ViolentCorpse
That depends. Spectacles are corrective lenses, meaning that they are designed to correct for aberrations introduced by imperfections in the eye. Sometimes they are slightly positive, sometimes negative, and sometimes they have little to no optical power, but correct for aberrations other than nearsightedness and farsightedness. A lens that corrects for astigmatism has some optical power, but only in one axis.

Sure there is. An object place around 15-25 mm from the eye will be at one focal length and the light will not form an image on the retina. When we say that an image isn't formed, we mean that the light leaving a lens/mirror is parallel to it itself instead of converging or diverging. However, you can still see an 'image' on your retina or on a camera lens, it's just extremely blurry.

I can't find a proper image, so we'll have to improvise. In the image below, imagine that the light is leaving an object at the focal point instead of converging on it. The light rays are refracted so that they emerge parallel to themselves. Now, imagine your retina is on the left. All of the light will fall on your retina within the diameter of that column of light, and none will be outside of it. Light emitted from a point just below the illustrated focal point will form its own column of parallel light which will mostly overlap the previous column except for a small sliver on the edge. So if you put your finger very close to your eye you will see a large, blurry blotch where most of the different columns of light from your finger are overlapping, but the edge of your finger will be an extended blur which you can see more distant objects through.

Thanks a ton, Drakkith! You are awesome!