Question

A converging lens (f = 11.7 cm) is located 26.6 cm to the left of a diverging lens (f = -5.42 cm). A postage stamp is placed 39.7 cm to the left of the converging lens. (a)Locate the final image of the stamp relative to the diverging lens. (b) Find the overall magnification.

Answer 1

For converging lens ::

f = 11.7 cm

d_o = distance of stamp = 39.7 cm

d_i = image distance

using the lens equation

1/d_o + 1/d_i = 1/f

1/39.7 + 1/d_i = 1/11.7

d_i = 16.6 cm to the right of converging lens

d = distance between two lenses = 26.6 cm

For diverging lens :

Image formed by converging lens acts as the object for diverging lens

so d'_o = object distance for diverging = 26.6 - 16.6 = 10 cm

f = - 5.42 cm

using the lens equation

1/d'_o + 1/d'_i = 1/f

1/10 + 1/d'_i = 1/(-5.42)

d_i' = - 3.5 cm to the left of diverging lens

b)

For converging lens

h_i = height of image

h_o = height of object

using the formula

h_i / h_o = - d_i/d_o

h_i / h_o = - 16.6/39.7

h_i = - 0.42 h_o          eq-1

for diverging lens

height of object = height of image formed by converging lens = h'_o = h_i = - 0.42 h_o

height of image = h'_i

using the equation

h'_i / h'_o = - d_i/d_o

h'_i / h'_o = - (-3.5/10)

h'_i / (-0.42 h_o) = 0.35

h'_i / h_o = - 0.147

m = - 0.147