Page 416 - Schaum's Outline of Theory and Problems of Applied Physics
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CHAP. 32]                                 LENSES                                      401



                  In general, an object that is the distance 2 f from a converging lens has a real, inverted image the same size as
              the object with an image distance equal to 2 f .


        SOLVED PROBLEM 32.10
              A key 6 cm long is 100 cm from a converging lens whose focal length is 40 cm. Find the location, size,
              and nature of the image.
                  Here p = 100 cm and f =+40 cm, so the image distance is
                                             pf    (100 cm)(40 cm)
                                        q =      =              = 66.7cm
                                           p − f   100 cm − 40 cm
              The image is real since q is positive (Fig. 32.8). The length of the key’s image is
                                              q           66.7cm
                                       h =−h    = (−6cm)         =−4cm

                                              p           100 cm
              The image is inverted (since h is negative) and is smaller than the object.

                  In general, an object that is farther than 2 f from a converging lens has a real, inverted image smaller than the
              object with an image distance between f and 2 f .

        SOLVED PROBLEM 32.11

              A diverging lens has a focal length of −2 ft. What are the location, size, and nature of the image formed
              by the lens when it is used to look at an object 12 ft away?
                  Here p = 12 ft and f =−2 ft, so the image distance is

                                             pf     (12 ft)(−2ft)
                                        q =      =             =−1.71 ft
                                            p − f  12 ft − (−2ft)
              A negative image distance signifies a virtual image. The magnification is

                                               q    (−1.71 ft)       1
                                         m =−    =−         = 0.143 =
                                               p      12 ft          7
              The image is erect (since m is positive) and one-seventh the size of the object.


        SOLVED PROBLEM 32.12
              A double-convex lens has a focal length of 6 cm. (a) How far from an insect 2 mm long should the lens
              be held in order to produce an erect image 5 mm long? (b) What is the image distance?

              (a) A double-convex lens is always converging, so the focal length of the lens is +6 cm. An erect image means a
                  positive magnification, which is

                                                     h    5mm
                                                 m =   =      = 2.5
                                                     h   2mm
                  Since m =−q/p, the image distance is q =−mp. We proceed by substituting q =−mp in the lens equation
                  and solving for p:
                                     1   1   1      1    1   1      m − 1   1
                                       =   =          −    =              =
                                     p   q   f      p   mp    f      mp     f

                                               m − 1           2.5 − 1
                                        p = f        = (+6cm)         = 3.6cm
                                                m                2.5
              (b)                          q =−mp =−(2.5)(3.6cm) =−9cm
                  The negative image distance signifies a virtual image (Fig. 32-9).
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