Question

A converging lens is placed at x 0, a distance d 11.5 cm to the left of a diverging lens as in the figure below (where F and Fo locate the focal points for the converging and the diverging lens, respectively). An object is located et x-1.90 cm to the left of the converging lens and the focal lengths of the converging and diverging lenses are 4.50 cm and -8.60 cm, respectively. 

(a) Determine the x-location in cm of the final image

(b) Determine its overall magnification

Answer 1

f=4.50cm f2-8.60cm I 4 11,50cm X=-5.44cm X--1-70 X0 x=11.5cm ~ 4.50x1.90 for converging lens object distance, 0 = -1.90cm Image distance, I = ? focal length f= 4.50cm using lens formula $ À Ó> * = $ * • -0.50 1.90 (1090-4.50) = -3.28 cm l.e., on the same side of object at x==3.29 cm magnification om, a 2/2 = -3.29 - 1173 for divenging lens, this image acts as object, Hence Object distance, u=-(11.50 +3.29)=-14.79 cm focal length, f = -8.60cm Image distance, sa fie - 23.39 =-5.44 cm magnification my = 2 = 5.44 = 0.37 x-location of final image x=-5.44 cm Overall magnification mam, m = 107340:37 m = 0.67 a 8.60x14.79 14079 2 6