This is one of my practice problems for my upcoming exam. I'm supposed to find the position of the image and the angular magnification.

A small insect is placed 6.35 cm from a 7.00 cm -focal-length lens.

I've calculated the position of the image to be -68 cm which is correct.

As for the angular magnification, isn't it just M = N/f = 25 cm/ 7 cm = 3.57.

But it's not the right answer.

  • $\begingroup$ Welcome to Physics SE. Are you using "eyepiece magnification"? If so, it is not the one you are looking for. Check your geometry to see why, and try using the distances to image and object instead . $\endgroup$ – udrv Nov 1 '16 at 2:12
  • $\begingroup$ Hi, I was suggested to try M = hᵢ / hₒ = -dᵢ / dₒ = 68.38/6.35 ~= 10.77 . I tried this, but it still wasn't the correct answer. $\endgroup$ – Valeria Nov 1 '16 at 2:22

The important factor is what is called the visual angle.
It is the angle subtended by the object or image at the eye.
This angle then determines the size of the image formed on the retina or photosensitive surface in a camera.

Suppose the object size is $d$ then the best you can do in terms of making the visual angle as large as possible is to place the object at the near point which is usually taken to be 25 cm.
This is the closest that you can put the object and still be able to focus on it.
If you try and find this distance yourself you most likely will find that your least distance of distinct vision is not 25 cm.
It might be less than 25 cm if you are young and more than 25 cm if you are old, so this is just an average sort of number.

For small angles, the visual angle with the object 25 cm away is $\frac {d}{25}$.

When looking through the lens the visual angle is $\frac{d}{6.35}$.

The ratio of these two visual angles is the angular magnification.

The largest angular magnification you can achieve with the lens is to put the object at such a distance from the lens that the virtual image is 25 cm from the lens.
The eye will then still be able to focus the light from the lens.
This then bring the object at the closest point to the lens whilst still forming an image which the eye can focus.


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