Question

A converging lens of focal length 8.080 cm is 20.0 cm to the left of a diverging lens of focal length -6.91 cm . A coin is placed 12.1 cm to the left of the converging lens.

di =   cm to the right of the diverging lens

Find the magnification of the coin's final image.

m=

Answer 1

for first lens, f1 = 8.08 cm

object distance, u1 = 12.1 cm

let v1 is the image distance.

1/u1 + 1/v1 = 1/f1

1/v1 = 1/f1 - 1/u1

1/v1 = 1/8.08 - 1/12.1

v1 = 24.3 cm

magnification, m1 = -v1/u1 = -24.3/12.1 = -2.01

for second lens, f2 = -6.91 cm

object distance , u2 = -24.3 + 20

= -4.30 cm

let v2 is the image distance.

use, 1/u2 + 1/v2 = 1/f2

1/v2 = 1/f2 - 1/u2

1/v2 = 1/(-6.91) - 1/(-4.3)

v2 = 11.4 cm <<<<<<<<------------------Answer

magnification, m2 = -v2/u2

= -11.4/(-4.3)

= 2.65

the magnification of the coin's final image, m = m1*m2

= -2.01*2.65

= -5.33 <<<<<<<<---------------------Answer