Question

A diverging lens with a focal length of 14 cm is placed 11 cm to the right of a converging lens with a focal length of 20 cm . An object is placed 41 cm to the left of the converging lens. Where will the final image be located? Express your answer using two significant figures. EVO ALQ * R O 2 ? d= cm to the left of the diverging lens Submit Request Answer Part B Where will the image be if the diverging lens is 36 cm from the converging lens? Express your answer using two significant figures. Find the image location relative to the diverging lens. O AO OD ? cm to the right of the diverging lens Submit Request Answer

Part A and B

Answer 1

A)
let f1 = 20 cm
u1 = 41 cm
let v1 is the image distance due to first lens.

use, 1/u1 + 1/v1 = 1/f1

1/v1 = 1/f1 - 1/u1

1/v1 = 1/20 - 1/41

v1 = 39.0 cm

let f2 = -14 cm

object distance for the second lens, u2 = -39 + 11

= -28 cm

let v2 is the image distance due to second lens.

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

1/v2 = 1/f2 - 1/u2

1/v2 = 1/(-14) - 1/(-28)

v2 = -28 cm

d = 28 cm <<<<<<<<<<<<--------------------Answer

B)

object distance for the second lens, u2 = -39 + 36

= -3 cm

let v2 is the image distance due to second lens.

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

1/v2 = 1/f2 - 1/u2

1/v2 = 1/(-14) - 1/(-3)

v2 = 3.82 cm

d = 3.82 cm <<<<<<<<<<<<--------------------Answer