I have several magnifiers/magnifying-glasses of varying quality with either nothing inscribed on them, or power ratings that I think are wrong.

How can the power of a magnifier be determined when no values are given?

Kyle Downs at quora.com states:

Magnification $M$ of a single lens is $M=\frac{f}{f-d}$

I don't know for sure if my magnifiers are "single-lenses", so that may complicate this. According to edmundoptics.com:

"Most high quality magnifiers use achromats [a positive simple lens cemented to a negative simple lens] to eliminate color fringing at the edge of objects."

Thus I imagine my higher-quality "jeweler's loupes" may in fact have multiple lenses.

As I understand, focal length can be found by moving a shining a light source and moving a lens so that the projected image of the source displays sharpest on the surface. The distance from surface to lens is the focal length.

Focal length diagram
(Image from wikipedia.org)

To experiment, I measured the focal length of one my loupes to be about 9mm, and the diameter of the lens about 16mm. Using the formula mentioned above, that would give me a magnification power of 9/(9-16) = -1.29x ...Which I feel cannot be right. It is much stronger and should be more around 20x or 30x. Plus, the image is not inverted, so the magnification should be a positive value, correct?

Clearly I am going about this wrong.


2 Answers 2


Having made a hash of my original answer which was pointed out by @Pieter, I have rewritten it.

In this experiment I simultaneously look at an object (black lines ruled 10 mm apart) which is 250 mm from the magnifier/eye and at the virtual image of mm square graph paper formed 250 mm from a magnifier/eye.
Fine adjustment was made by reducing the parallax between the black lines and the virtual image of the graph paper.
Although readings are best taken using an eye I have taken a photograph with a hand held iPhone to show what is seen.
When viewed with the eye the black lines and the virtual image were sharp but I could not achieve that with the iPhone (right hand) photograph.

enter image description here

The experimental arrangement is shown in the left hand diagram.

I used a x10 magnifier and by eye it was relatively easy estimate the magnification as being approximately 10.

There is a real difficulty measuring the short focal lengths of such lenses.

A method which is relatively easy to perform uses a plane mirror and finds the position of a virtual method by using no parallax.

  • $\begingroup$ It should say it on the other side. Maybe X8? Like here picclickimg.com/d/l400/pict/253390554186_/… $\endgroup$
    – user137289
    Commented Oct 12, 2020 at 0:02
  • $\begingroup$ The magnifying power of loupes (and of compound microscopes) is defined in relation to what one can observe with the naked eye at a distance of 25 cm. If you don't get x10 with that folding loupe you are not using it as intended: the loupe close to your eye, the object at the focal distance (2.5 cm) or a little bit closer. Magnification is something different. $\endgroup$
    – user137289
    Commented Oct 12, 2020 at 13:01
  • 1
    $\begingroup$ @Pieter Thanks for pointing out the error in my original answer. I have rewritten it. $\endgroup$
    – Farcher
    Commented Oct 12, 2020 at 23:18
  • $\begingroup$ Very nice photos of the experiment! $\endgroup$
    – user137289
    Commented Oct 12, 2020 at 23:33

The power of a magnifying glass is relative to the near-limit distance of the standard human eye (25 cm). The lens makes it possible to bring the object closer to the eye, so that it subtends a larger angle than at the near limit. The ratio of the (tangent of the) angles is proportional to the ratio of these distances.

Ray diagram

So if the focal length of a loupe is 10 mm, its M = 25.

  • $\begingroup$ Using 25cm gives: 250mm/(250mm-16mm) = 1.07x. Something is still not right. $\endgroup$
    – Bort
    Commented Oct 11, 2020 at 22:09
  • $\begingroup$ @Bort That formula is wrong, that is what I was trying to say. Use $M= 25\ {\rm cm}/f$ (or add 1 if you prefer). $\endgroup$
    – user137289
    Commented Oct 11, 2020 at 22:13

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