# Topic 38. The simple magnifying glass

In this section we describe how a single positive lens can be used as a simple magnifying glass.  Recall that the magnification of a lens is the ratio of the size of the image, si to the size of the object, so. As we have shown before (Topic 31. Camera Lenses, Part 3. The Effect of Focal Length), the magnification of a simple lens is the ratio of the distance from the lens to the image, di divided by the distance from the lens to the object, do:

The distance, di , between the eye-lens and the retina is fixed by the size and shape of the eye. Therefore, the magnification can be changed only by changing the distance between the object and the eye, do.  Since this distance is in the denominator of the fraction above, the magnification increases as an object is brought closer to the eye.  The maximum magnification will be realized when the object is brought to the near point. The magnification could be increased by bringing the object still closer, but the strength of the eye-lens can not be made great enough to bring the object into focus at these short distances. Since the distance between the eye-lens and the retina is about 20 mm and the near-point is about 25 cm, the maximum magnification of the eye is about 2/25= 0.08. (Note that this magnification describes the relationship between physical size of the image on the retina and the physical size of the object. All animals learn to relate the physical size of an image on the retina back to the physical size of the object that produced it.)

# There are two extreme situations: (1) the eye lens is configured with as short a focal length as possible so that it is still focused on the object at the near-point. If a person then puts an additional positive lens whose focal length is f (in cm) in front of the eye, and if the eye-lens remains set to its shortest possible focal length, then the magnification (relative to the magnification of the eye by itself as described above) is

In order to be in sharp focus in this situation, the object must be placed at a distance of

Since the denominator of the fraction is always greater than the numerator, the value of the fraction is always less than 1, so that the object distance is less than the focal length of the positive lens that has been added. For example, if the focal length of the additional lens is 5 cm, the magnification is 1+5=6, and the object will be in focus when it is about 4.2 cm in front of the eye.

The second situation is when the eye lens is configured to be as relaxed as possible so that it is focused on a very distant object. If a person then puts an additional positive lens whose focal length is f (in cm) in front of the eye, and if the eye-lens remains set to its longest possible focal length, then the magnification (relative to the magnification of the eye itself as described above) is

In order to be in sharp focus in this situation, the object must be located at a distance equal to the focal length of the additional lens. That is,

Since this distance is somewhat greater than the distance in the first situation, it should not be surprising that the magnification is somewhat smaller. Using the same additional lens as above, the magnification would be 5 and the object would be in clear focus when it was 5 cm in front of the eye.