Question

Constants | Periodic Table Part A A diverging lens with a focal length of -13 cm is placed 10 cm to the right of a converging lens with a focal length of 18 cm . An object is placed 37 cm to the left of the converging lens. Where will the final image be located? Express your answer using two significant figures. A2 Submit Previous Answers Request Answer Incorrect; Try Again; 5 attempts remaining Part B Where will the image be if the diverging lens is 33 cm from the converging lens? Express your answer using two significant figures. Find the image location relative to the diverging lens.

Answer 1

here,

for converging lens

focal length , f1 = 18 cm

object distance , do1 = 37 cm

let the image distance be di1

using the lens formula

1/f1 = 1/di1 + 1/do1

1/18 = 1/di1 + 1/37

solving for di1

di1 = 35.05 cm

for diverging lens

focal length , f2 = -13 cm

object distance , do2 = 10 - di1 = - 25.05 cm

let the image distance be di2

using the lens formula

1/f2 = 1/di2 + 1/do2

- 1/13 = 1/di2 - 1/ 25.05

solving for di2

di2 = - 27 cm

the final image is 27 cm behing the diverging lens

b)

for converging lens

focal length , f1 = 18 cm

object distance , do1 = 37 cm

let the image distance be di1

using the lens formula

1/f1 = 1/di1 + 1/do1

1/18 = 1/di1 + 1/37

solving for di1

di1 = 35.05 cm

for diverging lens

focal length , f2 = -13 cm

object distance , do2 = 33 - di1 = - 2.05 cm

let the image distance be di2

using the lens formula

1/f2 = 1/di2 + 1/do2

- 1/13 = 1/di2 - 1/2.05

solving for di2

di2 = 2.43 cm

the final image is 2.43 cm to the right of the diverging lens