An object is placed 10 cm in front of a lens. The image formed is real, inverted and of same size as the object. What is the focal length and nature of the lens?

Correct — A, 5 cm, converging. A real inverted image the same size as the object is the signature of an object sitting at twice the focal length. Its image then forms at 2f on the far side with magnification exactly −1. Since the object is at 10 cm, 2f = 10 cm and f = 5 cm. The lens formula agrees: with u = −10 cm and m = −1 the image distance is v = +10 cm, so 1/f = 1/v − 1/u = 1/10 + 1/10 = 1/5, giving f = 5 cm. Only a converging (convex) lens can produce a real image of a real object at all.

• (b)10 cm, diverging — A diverging lens produces a virtual, erect, diminished image of a real object at every position, so it can never give the real inverted image the stem describes. The 10 cm also repeats the object distance rather than halving it.
• (c)20 cm, converging — With f = 20 cm the object at 10 cm lies inside the focus, which gives a virtual, erect, magnified image — the magnifying-glass case, not a real one.
• (d)20 cm, diverging — This carries both faults at once: a diverging lens cannot form a real image, and 20 cm is not half the object distance.

For a thin lens, 1/f = 1/v − 1/u and magnification m = v/u. Four positions of the object matter: beyond 2f gives a real, inverted, diminished image; exactly at 2f gives real, inverted and equal in size; between f and 2f gives real, inverted and magnified; inside f gives virtual, erect and magnified. Only a convex lens runs through this sequence — a concave lens gives virtual, erect and diminished images throughout.

The phrase 'same size' is the whole question. It fixes |m| = 1, and for a lens that happens at one place only, the 2f point, so the focal length is half the object distance without any algebra. The word 'real' does the second half of the work by eliminating both diverging options before any arithmetic starts. A candidate who reaches for the formula first still gets there, but the two-step reasoning takes about ten seconds.

• For a thin lens, 1/f = 1/v − 1/u with distances measured from the optical centre and the incident direction taken positive.
• Magnification m = v/u; a negative m means a real, inverted image.
• An object at 2f gives a real, inverted image of equal size at 2f on the other side.
• A concave (diverging) lens always gives a virtual, erect, diminished image of a real object.
• Power in dioptres is 1/f in metres, so f = 5 cm corresponds to +20 D.

• Assuming a diverging lens can form a real image of a real object.
• Taking the focal length to equal the object distance instead of half of it.
• Dropping the minus sign on u and getting f = 10 cm from 1/f = 1/10 − 1/10, which gives an infinite focal length.

A standard image-characteristics item: the examiner gives the image description and asks for the lens, which is the reverse of the usual textbook drill.

• practice — not a real PYQ

  An object is placed 30 cm in front of a convex lens of focal length 15 cm. The image is

  • (a)virtual, erect and magnified
  • (b)real, inverted and of the same size
  • (c)real, inverted and magnified
  • (d)real, erect and diminished

  Answer(b) real, inverted and of the same size — the object sits exactly at 2f.

• practice — not a real PYQ

  The power of a converging lens of focal length 5 cm is

  • (a)+5 D
  • (b)+20 D
  • (c)−20 D
  • (d)+0.05 D

  Answer(b) +20 D — power is 1/f with f in metres, so 1/0.05 = 20 dioptres, positive for a converging lens.