# How to determine the power of a magnifier lens empirically when no values are given?

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.

(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?

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.

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.

• It should say it on the other side. Maybe X8? Like here picclickimg.com/d/l400/pict/253390554186_/…
– user137289
Commented Oct 12, 2020 at 0:02
• 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.
– user137289
Commented Oct 12, 2020 at 13:01
• @Pieter Thanks for pointing out the error in my original answer. I have rewritten it. Commented Oct 12, 2020 at 23:18
• Very nice photos of the experiment!
– 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.

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

• Using 25cm gives: 250mm/(250mm-16mm) = 1.07x. Something is still not right.
– Bort
Commented Oct 11, 2020 at 22:09
• @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).
– user137289
Commented Oct 11, 2020 at 22:13