Question

An object is placed 16.4 m infront of a converging lens of focal lenght f1=40 cm. On the other side of the lens, there is a diverging lens of focal lenght f2=4 cm, seperated from the first lens by a distance of 36.9 cm.

A) Where if the final image formed?

b) What is the magnification and what is the angular magnification and how does the angular magnigifcation compare to the telescipe formula?

C) Support your conclusion by drawing an appropriate ray diagram and describe the final image in words. Whether it is upright or inverted? Magnified or reduced?

Answer 1

a)

for the converging lens

f₁ = focal length = 40 cm

d_1o = object distance = 16.4 m = 1640 cm

d_1i = image distance = ?

Using the lens equation

1/d_1o + 1/d_1i = 1/f₁

1/1640 + 1/d_1i = 1/40

d_1i = 41 cm

m₁ = magnification = - d_1i /d_1o = - 41/1640 = - 0.025

for the diverging lens :

d = distance between the two lenses = 36.9 cm

d_o2 = object distance = 41 - 36.9 = 4.1 cm

d_i2 = image distance = ?

f₂ = focal length = - 4 cm

using the lens equation

1/d_o2 + 1/d_i2 = 1/f₂

1/4.1 + 1/d_i2 = 1/(-4)

d_i2 = - 2.025 cm

m₂ = magnification = - d_i2 /d_o2 = - (- 2.025)/4.1 = 0.494

final image is located at distance 2.025 cm from the diverging lens and is not in between the two lenses

b)

m = magnification = m₁ m₂ = (- 0.025) (0.494) = - 0.01235