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A car when viewed through a concave lens looks upright.
A car viewed through a concave or diverging lens looks upright. This is a case 3 image. (credit: Daniel Oines, Flickr)
Figure (a) shows an upright object placed at d sub o equals seven point five cm and in front of a concave lens of on its left side. Parallel ray 1 falls on the lens and gets refracted and dotted backwards to pass through point F on the left side. Figure (b) shows ray 2 going straight through the center of the lens.  Figure (c) combines both figures (a) and (b) and the dotted line and the solid line meet at a point on the left side of the lens forming a virtual image which is erect and diminished. Here h sub o is the height of the object above the axis and h sub i is the height of the image above the axis. The distance from the center to the image is d sub i equals 4.29 cm.
Ray tracing predicts the image location and size for a concave or diverging lens. Ray 1 enters parallel to the axis and is bent so that it appears to originate from the focal point. Ray 2 passes through the center of the lens without changing path. The two rays appear to come from a common point, locating the upright image. This is a case 3 image, which is closer to the lens than the object and smaller in height.

Image produced by a concave lens

Suppose an object such as a book page is held 7.50 cm from a concave lens of focal length –10.0 cm. Such a lens could be used in eyeglasses to correct pronounced nearsightedness. What magnification is produced?

Strategy and Concept

This example is identical to the preceding one, except that the focal length is negative for a concave or diverging lens. The method of solution is thus the same, but the results are different in important ways.

Solution

To find the magnification m size 12{m} {} , we must first find the image distance d i size 12{d rSub { size 8{i} } } {} using thin lens equation

1 d i = 1 f 1 d o ,

or its alternative rearrangement

d i = fd o d o f . size 12{d rSub { size 8{i} } = { { ital "fd" rSub { size 8{o} } } over {d rSub { size 8{o} } - f} } } {}

We are given that f = –10.0 cm and d o = 7 . 50 cm size 12{d rSub { size 8{o} } =7 "." "50"" cm"} {} . Entering these yields a value for 1/ d i size 12{d rSub { size 8{i} } } {} :

1 d i = 1 −10.0 cm 1 7 . 50 cm = 0 . 2333 cm . size 12{ { {1} over {d rSub { size 8{i} } } } = { {1} over { - "10" "." 0" cm"} } - { {1} over {7 "." "50"" cm"} } = { { - 0 "." "2333"} over {"cm"} } } {}

This must be inverted to find d i size 12{d rSub { size 8{i} } } {} :

d i = cm 0 . 2333 = 4 . 29 cm . size 12{d rSub { size 8{i} } = - { {"cm"} over {0 "." "2333"} } = - 4 "." "29"" cm"} {}

Or

d i = 7 . 5 10 7 . 5 10 = 75 / 17 . 5 = 4 . 29 cm .

Now the magnification equation can be used to find the magnification m size 12{m} {} , since both d i size 12{d rSub { size 8{i} } } {} and d o are known. Entering their values gives

m = d i d o = 4 . 29 cm 7 . 50 cm = 0 . 571 . size 12{m= - { {d rSub { size 8{i} } } over {d rSub { size 8{o} } } } = - { { - 4 "." "29"`"cm"} over {7 "." "50"`"cm"} } =0 "." "571"} {}

Discussion

A number of results in this example are true of all case 3 images, as well as being consistent with [link] . Magnification is positive (as predicted), meaning the image is upright. The magnification is also less than 1, meaning the image is smaller than the object—in this case, a little over half its size. The image distance is negative, meaning the image is on the same side of the lens as the object. (The image is virtual.) The image is closer to the lens than the object, since the image distance is smaller in magnitude than the object distance. The location of the image is not obvious when you look through a concave lens. In fact, since the image is smaller than the object, you may think it is farther away. But the image is closer than the object, a fact that is useful in correcting nearsightedness, as we shall see in a later section.

[link] summarizes the three types of images formed by single thin lenses. These are referred to as case 1, 2, and 3 images. Convex (converging) lenses can form either real or virtual images (cases 1 and 2, respectively), whereas concave (diverging) lenses can form only virtual images (always case 3). Real images are always inverted, but they can be either larger or smaller than the object. For example, a slide projector forms an image larger than the slide, whereas a camera makes an image smaller than the object being photographed. Virtual images are always upright and cannot be projected. Virtual images are larger than the object only in case 2, where a convex lens is used. The virtual image produced by a concave lens is always smaller than the object—a case 3 image. We can see and photograph virtual images only by using an additional lens to form a real image.

Practice Key Terms 8

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Source:  OpenStax, Physics 101. OpenStax CNX. Jan 07, 2013 Download for free at http://legacy.cnx.org/content/col11479/1.1
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