Question

An optical system consists of two thin lenses: +100. mm and -20.0 mm. They are separated by a distance of 50.0 mm. An object is placed 50.0 mm in front of the +100. mm lens. Determine:

a.   the final image location, and

b.   transverse magnification.

c.   Is the final image real or virtual? Why?

Answer 1

Given f = 100 mm

        f ' = -20 mm

Separation x = 50 mm

Let the focal length of the combination be F then ,

(1/F) =(1/f) + (1/f ') -(x/ff ' )

       = ( 1/100) -( 1/20) -[50 /(-20 x100)]

      = 0.01 -0.05 +0.025

     = -0.015

F = 66.66 mm

Object distance u = 50 mm

image distance v = ?

From this ( 1/ u ) + ( 1/ v ) = ( 1/ F )

              ( 1/v ) +(1/50) = ( 1/ -66.66)

                          ( 1/ v ) = -0.035

                                v = - 28.57 mm

(a).Transverse magnification m = v / u

                                             = -0.57

(c).Final image is virtual