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A convex lens of focal length 30 cm creates an image of height 2 cm from an object placed at infinity. A concave lens of focal length 20 cm is then placed coaxially between the convex lens and the image, at a distance of 26 cm from the convex lens. What is the new size of the final image?

1. 2.5 cm
2. 5.0 cm
3. 1.25 cm
4. None of these

Correct answer: 2.5 cm

Solution

The convex lens (f = 30 cm) creates an image at 30 cm from it for an object at infinity. The concave lens is placed 26 cm from the convex lens, so the image acts as a virtual object 4 cm to the right of the concave lens (u = +4 cm in sign convention where distances from concave lens: real object to left is negative, virtual object to right is positive). For concave lens: 1/v - 1/u = 1/f => 1/v - 1/4 = 1/(-20) => 1/v = 1/4 - 1/20 = 5/20 - 1/20 = 4/20 = 1/5. So v = 5 cm. Magnification by concave lens = v/u = 5/4. New image height = 2 * (5/4) = 2.5 cm.

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