An object of height 2.4 cm is placed 29 cm in front of a diverging lens of focal length 19 cm. Behind the diverging lens, and 11 cm from it, there is a converging lens of the same focal length. The distance between the lenses is 5.0 cm. Find the location and size of the final image.

122.735 behind converging lens ; 2.16

Explanation:

Given tgat:

Object distance, u = 29 cm

Image distance, v =

Focal length, f = - 19 (diverging lens)

Mirror formula :

1/u + 1/v = 1/f

1/29 + 1/v = - 1/19

1/v = - 1/19 - 1/29

1/v = −0.087114

v = −11.47916

v = -11.48

Second lens

Object distance :

u = 11.48 + 11 = 22.48 cm

1/v = 1/19 - 1/22.48

1/v = 0.0081475

v = 1 / 0.0081475

v = 122.735 cm

122.735 behind second lens

Magnification, m

m = m1 * m2

m = - v / u

Lens1 :

m1 = -11.48 / 29 = - 0.3958620

m2 = - 122.735 / 22.48 = - 5.4597419

Hence,

- 0.3958620 * - 5.4597419 = 2.16

the answer is a. at the equator

the answer is a. the yacht traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

im doing peak rn so sry if its wrong