Page 161 - NCERT Science Class 10 English Medium
P. 161

Magnification
                                     The magnification produced by a lens, similar to that for spherical
                                     mirrors, is defined as the ratio of the height of the image and the height
                                     of the object. Magnification is represented by the letter m. If h is the
                                     height of the object and h′ is the height of the image given by a lens, then
                                     the magnification produced by the lens is given by,

                                             Height of the Image  h′
                                         m =                     =                                     (9.9)
                                             Height of the object  h
                                         Magnification produced by a lens is also related to the object-distance
                                     u, and the image-distance v. This relationship is given by
                                         Magnification (m ) = h′/h = v/u                              (9.10)
                                         Example 9.3
                                         A concave lens has focal length of 15 cm. At what distance should
                                         the object from the lens be placed so that it forms an image at 10 cm
                                         from the lens? Also, find the magnification produced by the lens.
                                         Solution
                                         A concave lens always forms a virtual, erect image on the same side
                                         of the object.
                                         Image-distance v    = –10 cm;
                                         Focal length f      = –15 cm;
                                         Object-distance u   = ?

                                               1   1    1
                                         Since   −    =
                                               v   u    f
                                             1   1  1
                                         or,   =  –
                                             u   v  f
                                             1    1     1       1   1
                                               =     –      =  –  +
                                             u   –10  (–15 )   10 15
                                             1   − 3 +  2  1
                                               =       =
                                             u    30     − 30
                                         or,  u = – 30 cm
                                         Thus, the object-distance is 30 cm.
                                         Magnification m = v/u
                                                 −10cm    1
                                            m =          =  ;  + 0.33
                                                 − 30cm    3
                                         The positive sign shows that the image is erect and virtual. The image
                                         is one-third of the size of the object.

                                         Example 9.4
                                         A 2.0 cm tall object is placed perpendicular to the principal axis of a
                                         convex lens of focal length 10 cm. The distance of the object from the
                                         lens is 15 cm. Find the nature, position and size of the image. Also
                                         find its magnification.



               156                                                                                   Science


                                                           2024-25
   156   157   158   159   160   161   162   163   164   165   166