Question

A diverging lens (f=−12cm) is placed 20cm to the left of a converging lens (f=6cm). A small coin of width 2.5cm is placed 25cm to the left of the diverging lens.

Part A:What is the image distance di for the final image (the image formed by the rightmost lens)?

Part B:What is the width of the final image?

Part c: Is the final image real or virtual?

Answer 1

using lens formula for lens 1

1/f = 1/v - 1/u

- 1/ 12 = 1/v + 1/25

v = 8.108 cm

now the above image acts as object for lens 2

distance of object for lens 2

u' = 20 + 8.108 = 28.108 cm

using lens formula

1/6 = 1/v' + 1/28.108

v' = 7.628 cm

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b)

using magnification

h' / h = (v/u) (v'/u')

h' / 2.5 = (8.108/ 25)* ( 7.628 / 28.108)

h' = 0.22 cm

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c)

image is real and inverted

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