# Can you find lens magnification from focal length?

• B
If you know the focal length of a lens and use the lens equation 1/do + 1/di = 1/f by assigning some arbitrary do and solve for di, can you then find the magnification just from the focal length from -di/do?

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If you know the focal length of a lens and use the lens equation 1/do + 1/di = 1/f by assigning some arbitrary do and solve for di, can you then find the magnification just from the focal length from -di/do?
magnification is related to how large one can see the object when placed at say below the focal point towards the pole -to get a magnified image. a human eye can see things clearly at say least distance of distinct vision D; so some people can say that D and focal length can give an estimate of magnification m,
somewhere i have seen people using (D/f) as magnification but you have to relate it to the angles subtended by an object and its image at the pole of the lens so that their ratio can give you idea of m. try to make a ray diagram

sophiecentaur
Gold Member
When you buy a Loupe (Jeweller's eye lens) it has a Magnification figure stamped on it. I guess that makes assumptions about the accommodation that the user has, to focus close images.

jtbell
Mentor
I guess that makes assumptions about the accommodation that the user has, to focus close images.
The standard formula for angular magnification of a magnifying lens (loupe) assumes that the object is at the focal point and the image is therefore at infinity; and that when the user is looking at the object directly, without the lens, he places it 25 cm in front of his eye. $$M = \frac{25 cm}{f}$$

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• sophiecentaur
jtbell
Mentor
Just to avoid confusion...

After looking at the first post I think it's clear that the OP is referring to linear magnification whereas the rest of us are referring to angular magnification!

sophiecentaur
Gold Member
Just to avoid confusion...

After looking at the first post I think it's clear that the OP is referring to linear magnification whereas the rest of us are referring to angular magnification!
Now you have introduced something that I don't quite get. Are we just talking trigonometry here?

Andy Resnick
Just to avoid confusion...

After looking at the first post I think it's clear that the OP is referring to linear magnification whereas the rest of us are referring to angular magnification!
Much of the OP is unclear. A lens can image with a range of magnifications; real lenses are optimized to operate within a restricted range.

jtbell
Mentor
Now you have introduced something that I don't quite get. Are we just talking trigonometry here?
Linear magnification is the simple ratio of image size to object size, with a +/- sign depending on whether the image is upright or inverted. Via similar triangles on a ray diagram, this equals the ratio of image distance to object distance, with a sign flip if you're using the Gaussian convention for signs. Most textbooks write this as something like m = hi/ho = -di/do.

Angular magnification is the ratio of the angle subtended by the image as seen by the device's user, to the angle subtended by the object when viewed directly. Strictly speaking you have to use trig to get the angles from the object and image sizes and locations, but at the introductory level we always use the small-angle approximation θ ≅ tan θ ≅ sin θ, using whichever function is easier to use from the ray diagram.

Much of the OP is unclear.
I take it to mean, "given f and do, can you use the resulting di to find the (linear) magnification using -di/do?" I would say "yes", but the resulting m will be true only for that value of do, which fits with rest of your statement, about a "range of magnifications".

Fervent Freyja
Gold Member
Well, no, but take it further and use those distances to plug into the magnification formula jtbell gave above. You will have to assign an arbitrary height for the object, but it should give you magnification. Of course, if that was even what you were asking about...

I'm just saying that you know that you can place an object in front of the lens and that light from the object will pass through the lens and form an image, whether or not the image is real or virtual. So, given the focal length, you could just plug in some object distance and find the resulting image distance from the thin lens equation. Now that you have those two values, you have the magnification. Thus, you're getting magnification directly from focal length through the thin lens equation.

I only asked if this is possible because it feels like I'm misunderstanding something, as though you need more initial information to find magnification than just focal length, but it seems from the answers that I was correct. I was indeed talking about linear magnification, not angular, so I should have been more clear on that.

sophiecentaur
Gold Member
Linear magnification is the simple ratio of image size to object size, with a +/- sign depending on whether the image is upright or inverted. Via similar triangles on a ray diagram, this equals the ratio of image distance to object distance, with a sign flip if you're using the Gaussian convention for signs. Most textbooks write this as something like m = hi/ho = -di/do.

Angular magnification is the ratio of the angle subtended by the image as seen by the device's user, to the angle subtended by the object when viewed directly. Strictly speaking you have to use trig to get the angles from the object and image sizes and locations, but at the introductory level we always use the small-angle approximation θ ≅ tan θ ≅ sin θ, using whichever function is easier to use from the ray diagram.

I take it to mean, "given f and do, can you use the resulting di to find the (linear) magnification using -di/do?" I would say "yes", but the resulting m will be true only for that value of do, which fits with rest of your statement, about a "range of magnifications".
I see the distinction now and how the image and object linear heights relate with a real, inverted image. But it strikes me that, when you deliberately place a virtual image at infinity with a simple lens, the linear magnification would be infinite and the angular magnification would be a more realistic figure to quote. I have just been playing with a '10X' loupe and I can arrange to place the image over quite a range of distances (by the parallax method) but the angular 'size' doesn't change much, if at all.
Nothing is simple in this world!

Andy Resnick
<snip>Thus, you're getting magnification directly from focal length through the thin lens equation.
Not exactly- you also specified an object distance. You are free to vary the object distance and when you do, you will find that the magnification varies as well.

Andy Resnick
<snip>Angular magnification is the ratio of the angle subtended by the image as seen by the device's user, to the angle subtended by the object when viewed directly. Strictly speaking you have to use trig to get the angles from the object and image sizes and locations, but at the introductory level we always use the small-angle approximation θ ≅ tan θ ≅ sin θ, using whichever function is easier to use from the ray diagram.
.
That's not exactly right- your definition of angular magnification is correct, but that's not how the paraxial approximation is used. The rays associated with angular magnification are known as 'chief rays', while the rays associated with the paraxial approximation are 'marginal rays':

http://spie.org/publications/optipedia-pages/press-content/pm92/pm92_161_marginal_chief_rays

This is often glossed over in intro classes when discussing afocal systems like telescopes:

http://www.saburchill.com/physics/images5/211003030.jpg

here, α and β are the chief ray angles that are used to calculate angular magnification

Andy Resnick