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How is the angular magnification of a magnifying glass related with how much bigger the image appears to the user? Is it simply the same? If yes, why? (Let's assume that the object is placed in the focal plane for simplicity)

I was just playing around a bit with a magnifying glass with a focal length of $12\,$cm (so the angular magnification is about $2$). When I measured the size of the image on the surface of the lens, it was about $2.5$ times the size of the original object.

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It is not clear to me what you meant by " I measured the size of the image on the surface of the lens".

You calculated value of approximately 2 was done on the assumption that the object was in the focal plane of the convex lens of focal length $f = 12$ cm.
If the height of the size is $h$ the best the naked eye can do is to view the object at the least distance of distinct vision $D$ which is generally taken to be $25$ cm. The angular magnification is $\frac {25}{12} \approx 2$.

The highest magnification is obtained when the final virtual image is at the least distance of distinct vision which would require the object to be approximately $8$ cm from the lens. This would give an angular magnification of $\frac {25}{8} \approx 3$.

To measure the magnification you need to set your lens $12$ cm from a grid and observe the image of the grid through the lens which is formed at infinity .
At the same time you need to observe another grid which is 25 cm from the (other?) eye.
This is very difficult to do.

Much easier is setting up the lens so that the final image is at the near point and I have done a simple experiment which produced a surprisingly good result.

enter image description here

I found a hand magnifier whose focal length was approximately $5$ cm and set it up to be about $4$ cm from the lens so that the virtual image would be about 25 cm from the lens.
I then put another grid 25 cm from the lens as shown in the photograph.
What was pleasing was that the iPhone simultaneously brought into focus the grid viewed through the lens and the grid 25 cm below the lens.
Note that the grid 4 cm from the lens was out of focus and "bigger" than the grid 25 cm from the lens.
If I had used that as my direct view grid as the reference the magnification found would have been in error.

10 magnified small squares were equal to 63 unmagnified small squares which gave a magnification of approximately $6$ which is not bad when compared with the theoretical value of $\frac {25}{4} \approx 6$.

So perhaps it is worth having another go at measuring the magnification of your $12$ cm lens noting that it is not only the focal length but the optical configuration which determines the magnification?


Later

enter image description here

The magnification $M$ of a magnifying glass is defined as

$$M = \dfrac{\text{angle subtended by image of object when 25 cm from the lens}}{\text{angle subtended by object when 25 cm from the naked eye}} = \dfrac {\alpha '}{\alpha}$$

The HyperPhysics article Simple Magnifier gives some more theory.

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  • $\begingroup$ Thanks a lot for your answer, Farcher! Unfortunately, I'm having trouble to understand what you did. For example, you write "I found a hand magnifier [...] and set it up to be about 4 cm from the lens [...]". What is the "lens" here? Do you have two lenses or are you referring to the lens of your camera? It would me so much easier, if you could show me in person what you did ;-) $\endgroup$ – Marc May 23 '16 at 18:30
  • $\begingroup$ When I wrote "I measured the size of the image on the surface of the lens" I did the following: I placed a ruler in the focal plane of the magnifying lens, looked through the glass, placed a second ruler right on the glass and measured the apparent size of the first ruler there. I'm not sure if this makes sense, understanding how the linear magnification should be defined in this case is part of my confusion. $\endgroup$ – Marc May 23 '16 at 18:36
  • $\begingroup$ I have added some labels to my photograph. I used the camera so as to show what one would see if one looked down with the naked eye. I understood what you did but it is not correct. The magnification measured relative to the size of the object when viewed at 25 cm from the naked eye. $\endgroup$ – Farcher May 23 '16 at 20:12
  • $\begingroup$ Thanks for the clarification and the captions! I think I now understand what you did and why you did it. Yes, this makes sense to me and answers most of my initial question. However, I still feel a bit confused, and a few questions remain: 1) Is the linear magnification as measured by your method the same as the angular magnification? If yes, how could I derive this theoretically? 2) What I measured clearly is some sort of magnification. How is it related to more sensible definitions of the magnifications which we have discussed? It would seem strange to me, if it wasn't related at all. $\endgroup$ – Marc May 23 '16 at 23:32
  • $\begingroup$ 1. In situation shown above the linear magnification is the same as the angular magnification. When objects/images are a long way away (at infinity) then you must use the angular magnification. 2. I set up your arrangement and the size of the image of the scale seen through the lens seemed to vary as I moved the position of my eye. I think this was because I could not simultaneously focus on the image and on the reference scale looked at directly. $\endgroup$ – Farcher May 25 '16 at 7:42

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