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I have various magnifying glasses and I'm using them when I take macro photos with a phone or a camera. I want to group/label my magnifying glasses by their magnification power. And by magnification power I mean something like 10x.

To calculate their magnification power, I've read several Wikipedia pages, Lens (Optics) for calculating focal length, and Magnification for calculating magnification power, but I end up with wrong results. I must be doing something wrong.

For example, I have this biconvex magnifying glass

enter image description here

Some of its properties

diameter = 6 cm
thickness at the edge = 3 mm
thickness at the center = 7 mm
radius of the curvature = 225 mm (1)

I figured the focal length of magnifying glass by experimenting with a light source, but I want to calculate it on a paper.

How can I calculate the focal length of a magnifying glass and its magnification power?

For example, in this page Amazon - SE Folding Pocket Magnifier, it says 10x for the magnifier.

How do the companies that produce magnifying glasses calculate it?

I'd appreciate if you could provide an example calculation in your answer, preferably with the values I've given.


(1) I didn't calculate it, I just drew the magnifying glass in Illustrator with its actual values
Index of refraction doesn't matter. The result doesn't have to be 100% correct. 1.52 can be used as the index.

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    $\begingroup$ One aspect of the lens that you need to consider is the material it is made with. The material influences the refractive index, which influences the magnification. As far as I know, opticians measure the lens to determine the magnification. $\endgroup$ – LDC3 Aug 3 '14 at 18:22
  • $\begingroup$ What is wrong with measuring? $\endgroup$ – Bernhard Aug 3 '14 at 18:28
  • $\begingroup$ Yes, I'm aware of that. But the result doesn't have to be 100% accurate. I suppose 1.52 can be used as an average for glasses. $\endgroup$ – akinuri Aug 3 '14 at 18:29
  • $\begingroup$ Could you please be more specific where your calculation following the Wikipedia got stuck? $\endgroup$ – Luboš Motl Aug 4 '14 at 9:04
  • $\begingroup$ @LubošMotl I couldn't apply the values I have with the given equations. I mean I end up with wrong values. I must be doing it wrong, or missing something. $\endgroup$ – akinuri Aug 4 '14 at 12:10
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A lens does not have have one specific magnification, it depends on the positioning of the lens. When neglecting aberrations, the workings of a lens can be simplified with the following equation,

$$ \frac{1}{f} = \frac{1}{v} + \frac{1}{b}, $$

where $f$ is the focal length of the lens, $v$ is the distance from the object to the lens and $b$ the distance from the lens to the image of the object. This is demonstrated in the image below, including three principal rays (these only apply for thin lenses).

schematic representation of the image formation of a lens

Here $\text{F}_1$ and $\text{F}_2$ are the two focal points of the lens, with $f_1$ and $f_2$ as their respective focal lengths (these are often equal to each other, which is also assumed in the first equation).

The resulting magnification, $M$, will be equal to the ratio between $h_1$ and $h_2$, which when expressed in terms of $v$ and $b$ looks as follows,

$$ M = \frac{v}{b}. $$

When you have a lens with a given focal length then you have two equations with three unknown. So, when you want to calculate the magnification you would not have a unique solution. However manufactures probably want to add a label to their lenses which a layman can understand. For this they probably will use eyepiece magnification,

$$ M_e = \frac{250\ mm}{f} $$

where the numerator is equal to the least distance of distinct vision, which is roughly 250 mm for a human with normal vision.

However if you do not know the focal length you can use the following equation,

$$ \frac{1}{f} = (n - 1) \left[\frac{1}{R_1} - \frac{1}{R_2} + \frac{(n - 1) d}{n R_1 R_2}\right], $$

where $n$ is the refractive index of the lens material, $d$ the thickness of the lens, $R_1$ and $R_2$ the radius of curvature of the two sides of the lens.

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  • $\begingroup$ Thank you, this visual was most likely very helpful. However, I don't think he has a sphereometer, and he only lists the radius of curvature for 1 side. In addition, it was lightly insinuated that he had a bit of trouble with the mathematical aspects of the problem. Also, it is not wise to assume that the lens is made of a uniform material, so the index of refraction on either side could throw off the calculation of lens maker's equation. It's best if he finds the focal length through the relationship M = (f/(f-d0)), and testing it for either side of the lens. $\endgroup$ – Gödel Aug 5 '14 at 22:31
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I believe that the difference between the radius of curvature and the thickness is great enough to use the thin lens equation [if not I have another formula at the end that will work with this method].

Let me know how it turns out:

  1. Take a flashlight and tape a cross to it(make sure you don't let any light through the taped section, this is very important).

  2. Make an optical setup aimed at the wall so that the flashlight is static and exactly 90 degrees to the center of your lens (make sure that you know how close your lens is to the wall!).

  3. Adjust your flashlight until the cross forms a clear image on the wall, and measure that distance with a meter stick (include an uncertainty if you wish).
  4. On a piece of paper write how far the flashlight was from the lens, and how far the lens was from the wall.
  5. divide the flashlight cross' distance the lens into (the number 1), then divide the lens distance to the wall into (the number 1).
  6. Add those two numbers together (assuming you used cm, multiply this by 100 and you have the power of your lens in inverse meters, or dioptres). *[in "maths" terms, take the inverses separately and combine them].
  7. divide that into (the number 1) as well(before multiplying it by 100) and now you have the focal length
  8. Repeat this process with at least 3 more distances to be certain. ***In addition, measure the cross' height on the wall, and divide that by the actual height of the cross. That will be the magnification (it will vary by object distance).

At the very least, even if the lens isn't thin enough to utilize the thin lens criterion, you will be able to calculate a magnification, by dividing the image's distance to the lens into the cross' distance to the lens (unless you used the height method I gave you first). After you have the magnification, you can use the relationship to calculate the focal length by the relationship: Magnification = (focal length)/(focal length - cross' distance from the lens). You have 1 equation and 1 unknown now. I am curious as to what you measure the focal length to be with each method. If you get different results, the lens was likely too thick to use the thin lens equation, however, you will still be able to calculate the focal length no matter what, and can therefore calculate the power by dividing that into 1 and multiplying it by 100.

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protected by Qmechanic Jun 28 '17 at 16:38

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