Question

When an object is placed at the proper distance to the left of a converging lens, the image is focused on a screen 30cm to the right of the lens. A diverging lens with focal length f=-24cm is now placed 15cm to the right of the converging lens.

How far and in which direction do we need to move the screen to get a sharp image?

How much larger is the new image?

Answer 1

The image of the converging lens become the object for the diverging lens.
The object is at a distance 15 cm in front of the diverging lens.
u = 15 cm
Focal length of the diverging lens, f = - 24 cm              [Negative sign is according to convention]
Consider v as the image distance.

Using the formula, 1/f = 1/u + 1/v
1/v = 1/f - 1/u
= - 1/24 - 1/15
= - 0.1083
v = - 9.23 cm
Negative sign indicates that the image is on the same side where the object is.

a)
Initially, the screen was at a distance of 15 cm in front of the diverging lens.
The screed must be moved to the left a distance of (15 - 9.23) = 5.77 cm

b)
Magnification = - (v/u)
= - (- 9.23/15)
= 0.62
The new image is now 0.62 times the previous image.