CDS (II) 2020 General Knowledge · Question 102 of 120

Correct — C, inverted and magnified. Put the numbers into the lens formula with the usual sign convention: the object is on the incoming side, so u = −40 cm, and the lens converges, so f = +25 cm. From 1/v − 1/u = 1/f, 1/v = 1/25 + 1/(−40) = 8/200 − 5/200 = 3/200, giving v = +200/3 ≈ 66·7 cm. A positive v means the image forms on the far side of the lens and is real, which fits the stem's statement that it is caught on a screen. The magnification is m = v/u = 66·7/(−40) ≈ −1·67. The minus sign says inverted, and the size 1·67 says the image is larger than the object. The qualitative check agrees: 40 cm lies between f and 2f — that is, between 25 cm and 50 cm — and an object placed between the focus and twice the focal length of a convex lens always gives a real, inverted, enlarged image beyond 2f on the other side.

• (a)erect and magnified — This is what a convex lens gives when the object is nearer than the focus — the magnifying-glass case, where the image is virtual. Here the object at 40 cm is well outside the 25 cm focus, so the image is real and therefore inverted.
• (b)erect and smaller — A converging lens never produces a smaller erect image; that is the signature of a diverging lens, which always gives a virtual, erect, diminished image whatever the object distance.
• (d)inverted and smaller — Right on orientation, wrong on size. A real inverted image is smaller only when the object lies beyond 2f — beyond 50 cm here. At 40 cm the object is inside 2f, so the image is enlarged. Missing that the object sits between f and 2f is the whole cost of this option.

For a thin converging lens the character of the image is governed entirely by where the object sits relative to f and 2f. Beyond 2f the image is real, inverted and diminished, formed between f and 2f. At exactly 2f it is real, inverted and the same size, formed at 2f. Between f and 2f it is real, inverted and enlarged, formed beyond 2f. At the focus the emergent rays are parallel and the image is at infinity. Inside the focus the image is virtual, erect and enlarged — the magnifying glass.

Two routes reach the answer and it is worth being able to run both. The arithmetic route is the lens formula plus m = v/u, which delivers the sign and the size together and is the only route that works when the paper asks for a number. The reasoning route is to compare the object distance with f and 2f, which here is a single comparison: 25 < 40 < 50, so the object is between f and 2f and the image must be real, inverted and enlarged. A useful reliability check is that any image received on a screen is real, and every real image formed by a single lens or mirror is inverted — so the words 'obtained in the screen' in the stem already eliminate both erect options before any calculation. The bank marks this question as carrying an image, but the printed page holds no figure — the item is purely numerical and the stem supplies every quantity needed.

• Lens formula: 1/v − 1/u = 1/f, with distances measured from the optical centre and the incoming direction taken as negative for the object.
• Magnification m = v/u; a negative m means an inverted image and |m| > 1 means an enlarged one.
• With u = −40 cm and f = +25 cm the image distance is v = +200/3 ≈ 66·7 cm and m ≈ −1·67.
• An object between f and 2f of a convex lens gives a real, inverted, enlarged image formed beyond 2f.
• Every image that can be caught on a screen is real, and a real image formed by a single lens is always inverted.

• Comparing the object distance with f only and forgetting to compare it with 2f, which is what decides enlarged against diminished.
• Dropping the minus sign on u and getting a negative v that suggests a virtual image.
• Treating a real image as possibly erect; a single lens gives real images only inverted.

Either numerically, as here, or as a position item — where must the object be placed to get a real and enlarged image — or as a magnification-sign item.

• practice — not a real PYQ

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

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

  Answer(b) real, inverted and of the same size — the object is exactly at 2f, the one position where a convex lens returns an image equal in size to the object, formed at 2f on the other side.

• practice — not a real PYQ

  The magnification produced by a lens is −0·5. The image is

  • (a)erect and enlarged
  • (b)erect and diminished
  • (c)inverted and enlarged
  • (d)inverted and diminished

  Answer(d) inverted and diminished — the negative sign marks inversion and the magnitude of 0·5 marks an image half the size of the object.