Question

Two thin lenses with a focal length of magnitude 12.0 cm, the first diverging and the second converging, are located 9.00 cm apart. An object 2.50 mm tall is placed 20.0 cm to the left of the first (diverging) lens. (a) How far from this first lens is the final image formed? (b) Is the final image real or virtual? (c) What is the height of the final image? Is it noninverted or inverted?

Answer 1

(a)

For diverging lens :

f = focal length = - 12 cm

d_o = object distance = 20 cm

h_o = object height = 2.50 mm = 0.25 cm

h_i = image height

d_i = image distance

using the lens equation

1/d_i + 1/d_o = 1/f

1/d_i + 1/20 = 1/(-12)

d_i = - 7.5 cm on the same side as the original object.

magnification is given as

h_i/h_o = - d_i/d_o

h_i/0.25 = - (- 7.5)/20

h_i = 0.09375 cm

This image behaves as object for the converging lens

For converging lens :

f = focal length = 12 cm

d_o = object distance = 9 + 7.5 = 16.5 cm

d_i = image distance

h_o = object height = 0.09375 cm

h_i = image height

using the lens equation

1/d_i + 1/d_o = 1/f

1/d_i + 1/16.5 = 1/(12)

d_i = 44 cm to the right of converging lens

D = distance from the diverging lens = 44 + 9 = 53 cm

b)

The final image is real since the object distance for converging lens is greater than its focal length.

c)

For converging lens :

magnification is given as

h_i/h_o = - d_i/d_o

h_i/0.09375 = - 44/16.5

h_i = - 0.25 cm

the height of the image is 0.25 cm and it is inverted