Question

An engraver uses a magnifying glass (f = 7.20 cm) to examine some work, as in the drawing. The image he sees is located 28.0 cm from his eye, which is his near point. (a) What is the distance between the work and the magnifying glass? (b) What is the angular magnification of the magnifying glass?

Answer 1

Given :

          Focal length ( f) = 7.2 cm

            Image distance ( v) = - 28 cm since image is virtual

          We know that :

                  1 / f   = 1/ u + 1 / v

                  1/ u   =   1/ f - 1/ v

                           = 1 /7.2 - 1/ (-28)

          ∴the distance between the work and the magnifying glass is ( u) = 5.73 cm

( b)

          Angular magnification is :

                        M = ( 1/ f - 1/v ) N

                              = ( 1/ 7.2 + 1/ 28 ) 28

                              = 4.89