An object placed 4 cm to the left of a lens with positive focal length f₁= +2 cm produces an image 4 cm to the right of the lens. A diverging lens, focal length f₂, is now placed at the focal point of the first lens, as shown, and produces a final image 6 cm to the right of lens f₂. The focal length f₂ is

Question

Quizplus · Accepted Answer

Using the thin lens formula, we can find the image distance for the first lens:
1/fS1U1B0 = 1/S1U1B1 - 1/S1U1B0
1/fS1U1B0 = 1/4 - 1/(-2)
fS1U1B0 = 4/3 cm (positive because it is a converging lens)

Now, we can use the diverging lens as a virtual object distance for the second lens (since it forms a virtual image for the first lens):
1/fS1U1B12S1U1B0 = 1/S1U1B2 - 1/fS1U1B0
1/fS1U1B12S1U1B0 = 1/6 + 1/4/3
fS1U1B12S1U1B0 = -3 cm (negative because it is a diverging lens)

Therefore, the focal length of the diverging lens is -3.0 cm, choice (B).

An Object Placed 4 Cm to the Left of a Lens