Question

Two lenses, one converging with focal length 21.0 cm and one diverging with focal length − 12.0 cm , are placed 25.0 cm apart. An object is placed 60.0 cm in front of the converging lens.

Part A

Find the final image distance from second lens. Follow the sign conventions.

Express your answer to two significant figures and

Part B

Determine the magnification of the final image formed. Follow the sign conventions.

Express your answer using two significant figures

include the appropriate units.

Answer 1

Image distance is given by
1/d_i + 1/d_o = 1/f

For first image formed by converging lens, we get
1/d_i + 1/60 = 1/21
d_i = 32.3
+ve sign indicates real image
magnification of first lens M1 = - d_o/d_i = -0.54
-ve sign indicates inverted image

This first image is object for second lens at distance of 25 cm.
d_o for this lens is -7.3 ( -ve because first image is formed beyond second lens, hence a virtual object)

so for second lens
1/d_i - 1/7.3 = -1/12
d_i = 18.63
hence final image is 19 cm ( two significant figures) from diverging lens. It's real image as d_i is +ve)

Magnification M2 = - (18.63/-7.3) = 2.55

Total magnification M = M1 x M2 = -1.4
hence an inverted image