Question

Two identical diverging lenses are separated by 15 cm. The focal length of each lens is -8.0 cm. An object is located 4.0 cm to the left of the lens that is on the left. Determine the final image distance relative to the lens on the right.

Answer 1

Light leaves the object and diverges over a distance of 4 cm to reach the 1st lens, so the vergence of the wavefronts entering the first lens is

V1 = - 1/(4 cm)

The vergence exiting the first lens is the entrance vergence plus the power of the lens:

V2 = V1+P1 = -1/(4cm) + 1/(-8cm) = -3/(8cm)

Since the vergence is negative the light is diverging, so this lens creates a virtual image upstream (left of it) a distance 1/|V2| away, or 8/3 cm to the left of the first lens.

That virtual image acts like an object for the 2nd lens. V3, the vergence of the wavefronts entering the 2nd lens is given by

V3 = - 1/(15cm + 8/3 cm) since the wavefronts diverge from the location of the intermediate image formed by the first lens. Simplifying,

V3 =.0566cm)

The light exiting the 2nd lens has vergence

V4 = V3 + P2 = 0.566cm + 1/(-8cm) = -0.441/cm.

This means that the light exiting the second lens looks like it originates a distance of magnitude 1/|V4| upstream (to the left) of the 2nd lens. So the image is virtual and a distance of 1/0.441cm^-1 = 2.267 cm to the left of the right hand lens.

Answer 2

1/p + 1/q = 1/f
1/4.0 + 1/q = 1/-8.0
q = -2.667 cm

Now use the first image as the second object. To find its distancefrom the second lens,
15 + 2.667 = 17.667

1/p + 1/q = 1/f
1/17.667 + 1/q = 1/-8.0
q = -5.506 cm

So the final image is 5.506 cm to the right of the lens on theright