Question

1.) The distance between an object and its image formed by a diverging lens is 46.1 cm. The focal length of the lens is -237.3 cm.

(a) Find the image distance.
cm

(b) Find the object distance.
cm

2.) An object is 19 cm in front of a diverging lens that has a focal length of -11 cm. How far in front of the lens should the object be placed so that the size of its image is reduced by a factor of 3.0?
  m

Answer 1

dot di = 46-1cm using thin lens equation do + di = $ ugl-dit di- - 23703 46.1 di (46-1-d;) - 23763 d-46-ldi - 10939.53=0 By sobing Quadratic equations di = (13015, -84.05197) cm we have discarded the positive root, because we know that di must be negative for the vistual image di =(-84cm b) do = 46c1cm-f-sucm) =113001 cm) 3) magnification m) = -di. m2 = m, a do do doz = 3diz (do) 2 - - T - - 1 - 1 dir = -6.967 cm de doe-doef de fidem dez-Zuf der Vor) > doz-f= 3f (* :) dog = 4 [()+) = (-1) [error doz = [79 cm