A converging lens of focal length 10cm forms an erect image three times the size of the real object. Calculate the distance between the image and the object.

Analyze the question: A converging lens gives an erect image that is 3 times the size of a real object. An erect image from a converging lens must be virtual, so the object is placed inside the focal length (this is the magnifying glass case).

Concept:

• Lens formula: 1/f = 1/u + 1/v (real-is-positive convention)
• Magnification: m = v/u. For a virtual image, v is negative, so |v| = 3u.

Solution:

Using the real-is-positive convention, with f = +10 cm and v = −3u (the image is virtual, on the same side as the object):

1/f = 1/u + 1/v 1/10 = 1/u − 1/(3u) 1/10 = (3 − 1)/(3u) = 2/(3u)

So 3u = 20, giving u = 20/3 ≈ 6.67 cm.

The image distance is |v| = 3u = 20 cm (virtual).

Distance between image and object: Both lie on the same side of the lens, so the separation is the difference:

20 − 6.67 = 13.33 cm ≈ 13.3 cm

Check: m = v/u = 20/6.67 = 3 ✓. Also, u < f (6.67 < 10), which is consistent with an erect, virtual image.

Answer: C (13.3 cm).

Tip: When a question says "erect image" for a converging lens, remember that it is virtual. Then set v = −m·u and substitute into the lens formula. The object–image distance is |v| − u, because they are on the same side. Option A (6.6 cm) is the trap, since it is just the object distance.