A 28cmtall object is 22 cm to the left of a lens with a foca

A 2.8-cm-tall object is 22 cm to the left of a lens with a focal length of 11 cm. A second lens with a focal length of 5.0 cm is 32 cm to the right of the first lens. Calculate the distance between the final image and the second lens. Express your answer to two significant figures and include the appropriate units. d = 10 cm Calculate the image height. Express your answer to two significant figures and include the appropriate units.

Solution

Part A:

First find the location and size of the image created by the first lens -

use the universal lense formula -

1/p1 +1/q1 = 1/f1 where p1 is object distance = 22 cm

and, q1=image distance

f1= focal length = 11 cm

Therefore, we have

1/22 + 1/q1 = 1/11
=> 1/q1 = 1/11 - 1/22 = 1/22

So, q1 = 22 cm

this means the image is 22 cm to the right of the first lens.

the magnification of this image is m1 = -q1/p1 = -(22)/22 = -1.0

Again,
We find the final image, which is created by the other lens;
the object for this lens is the image created above, and this image is (32 - 22) = 10 cm to the right of the second lens, and we have

1/10 + 1/q2 = 1/5

1/q2 = 1/5 - 1/10 = 0.10

=> q2 = 10 cm

this means the final image is 10 cm to the RIGHT of the second lens

Part B:

the magnification of the image is

m2=-10/10 = -1,

so the total magnification is m=m1xm2= (-1)x(-1) = 1

so the final image is the same as the original object, i.e., 2.8 cm