Question

A converging lens with a focal length of 4.2 cm is located 20.7 cm to the left of a diverging lens having a focal length of -11.5 cm. If an object is located 9.2 cm to the left of the converging lens, locate and describe completely the final image formed by the diverging lens. Where is the image located as measured from the diverging lens? Submit Answer Tries 0/10 What is the magnification? Submit Answer Tries 0/10 Also determine, with respect to the original object whether the image is real or virtual, upright or inverted? Check all statements that are true. Select all that are True. The image is virtual The image is inverted The image is real The image is upright Submit Anchor Tries 0110

Answer 1

f = 4.20m I am wsing cartesian Sign convention. f = -11.5cm 5 20.7cm Using lens equation for the first lens. I I I =) !== + Here f = 4.20m 4,- - 9.2cm - V - 4.2 9.2 4.2x9.2 4.249.2 V = 7.728cm 42 = -(2007-7.728 ) 2 = -12.972 Similarly for dircoging tons (f = -11.5 cm) I - + 1 Here I e un - I t 1 a 1 -12.972-11.5 -11.5" 12.972 2 11.5X12.572 = 24.412 E) K = - 1.5X12.972 =-60958 24.472 11.5 12.972 1 - 6.10cm to the left of the diverging lens. So final image is [6.10.com magnification mamm= h. 2 m = 7.728 -610 =-0.395 - 9.2 - 12.972 m= -0.395 |ml=0.395 and since & is negative mis negative so so final image is virtual The image is inverted.