Question

A converging lens of focal length 8.050 cm is 20.4 cm to the left of a diverging lens of focal length -6.64 cm . A coin is placed 12.2 cm to the left of the converging lens.

1. Find the location of the coin's final image.

a. to the left of the converging lens

b. between the lenses

c. to the right of the diverging lens

2. Find the magnification of the coin's final image.

Express your answer using two significant figures.

m =

Answer 1

here,

for converging lens

the object distance , do1 = 12.2 cm

focal length , f1 = 8.05 cm

let the image distance be di1

using the lens formula

1/f1 = 1/di1 + 1/do1

1/8.05 = 1/di1 + 1/12.2

solving for di1

di1 = 23.7 cm

for diverging lens

object distance , do2 = 20.4 - di1 = - 3.27 cm

focal length , f2 = - 6.64 cm

let the image distance be di2

using the lens formula

1/f2 = 1/di2 + 1/do2

- 1/6.6 = 1/di2 - 1/3.27

solving for di2

fi2 = 6.48 cm

a)

the location of final image is c)to the right of diverging lens

b)

the magnification , m = (di1/do1) *(di2/do2)

m = ( 23.7/12.2) *(6.48 /(-3.27))

m = - 3.85