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By the end of this section, you will be able to:
  • Understand the optics of a simple magnifier
  • Characterize the image created by a simple magnifier

The apparent size of an object perceived by the eye depends on the angle the object subtends from the eye. As shown in [link] , the object at A subtends a larger angle from the eye than when it is position at point B . Thus, the object at A forms a larger image on the retina (see O A ) than when it is positioned at B (see O B ). Thus, objects that subtend large angles from the eye appear larger because they form larger images on the retina.

Two objects of the same size are shown in front of an eye. Object A is closer to the eye and forms an angle theta 2 with the optical axis. Object B is farther away and forms an angle theta 1 with the optical axis. Inside the eye, the rays strike the retina. Ray B prime is closer to the optical axis than ray A prime.
Size perceived by an eye is determined by the angle subtended by the object. An image formed on the retina by an object at A is larger than an image formed on the retina by the same object positioned at B (compared image heights O A to O B ).

We have seen that, when an object is placed within a focal length of a convex lens, its image is virtual, upright, and larger than the object (see part (b) of [link] ). Thus, when such an image produced by a convex lens serves as the object for the eye, as shown in [link] , the image on the retina is enlarged, because the image produced by the lens subtends a larger angle in the eye than does the object. A convex lens used for this purpose is called a magnifying glass or a simple magnifier .

Figure a shows an object with height h 0 in front of an eye, at the near point. An image that is smaller than the object is formed on the retina. Figure b shows a bi-convex lens between the eye and the object. Rays from the object go through this and enter the eye to form a larger image on the retina. The back extensions of the rays deviated by the lens converge behind the object to form an image that is larger than the object. The distance of this image from the lens is d subscript i and that of the object from the lens is d subscript o. The distance of the lens from the eye is l. The distance of the image from the eye is L. The height of the image is h subscript i.
The simple magnifier is a convex lens used to produce an enlarged image of an object on the retina. (a) With no convex lens, the object subtends an angle θ object from the eye. (b) With the convex lens in place, the image produced by the convex lens subtends an angle θ image from the eye, with θ image > θ object . Thus, the image on the retina is larger with the convex lens in place.

To account for the magnification of a magnifying lens, we compare the angle subtended by the image (created by the lens) with the angle subtended by the object (viewed with no lens), as shown in [link] . We assume that the object is situated at the near point of the eye, because this is the object distance at which the unaided eye can form the largest image on the retina. We will compare the magnified images created by a lens with this maximum image size for the unaided eye. The magnification of an image when observed by the eye is the angular magnification     M , which is defined by the ratio of the angle θ image subtended by the image to the angle θ object subtended by the object:

M = θ image θ object .

Consider the situation shown in [link] . The magnifying lens is held a distance from the eye, and the image produced by the magnifier forms a distance L from the eye. We want to calculate the angular magnification for any arbitrary L and . In the small-angle approximation, the angular size θ image of the image is h i / L . The angular size θ object of the object at the near point is θ object = h o / 25 cm . The angular magnification is then

M = θ image θ object = h i ( 25 cm ) L h o .

Using [link] for linear magnification

m = d i d o = h i h o

and the thin-lens equation

1 d o + 1 d i = 1 f

in [link] , we arrive at the following expression for the angular magnification of a magnifying lens:

M = ( d i d o ) ( 25 cm L ) = d i ( 1 f 1 d i ) ( 25 cm L ) = ( 1 d i f ) ( 25 cm L )

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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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