CDS (I) 2022 General Knowledge · Question 23 of 120

Correct — C, 30 cm. A convex lens gives a real image the same size as the object at exactly one object distance, and that is twice the focal length. With f = 15 cm, twice the focal length is 30 cm. The algebra confirms it in a line. Same size and real means the magnification is −1, so v = −u; substituting into 1/v − 1/u = 1/f with u = −30 cm gives 1/30 + 1/30 = 2/30 = 1/15, which is 1/f. The image forms 30 cm on the far side of the lens, real, inverted and equal in height to the object. This arrangement is the one point on the whole object–image curve where object distance and image distance are equal, and it is the position a photocopier lens is set at when a copy has to come out the same size as the original.

• (a)15 cm — The focal length itself. An object placed exactly at the focus gives no image at any finite distance — the rays leave the lens parallel and meet only at infinity.
• (b)10 cm — Inside the focal length, which is magnifying-glass territory. From there the lens produces a virtual, erect and enlarged image on the same side as the object, so there is no real image at all, let alone one of the same size.
• (d)40 cm — Beyond twice the focal length. That gives a real image, but a diminished one, formed somewhere between the focus and twice the focus on the far side. Only at exactly 30 cm does the image match the object in size.

The behaviour of a convex lens as the object is brought in from far away follows a fixed sequence. Beyond twice the focal length the image is real, inverted and diminished, and it lies between f and 2f on the other side. At exactly twice the focal length it is real, inverted and the same size, at 2f on the other side. Between f and 2f it is real, inverted and enlarged, lying beyond 2f. At the focus the image runs off to infinity. Inside the focus the image becomes virtual, erect and enlarged, on the same side as the object. The thin-lens formula 1/v − 1/u = 1/f generates the whole table.

The fastest route is to recognise the phrase 'real image of the same size' as naming one specific row of that table, and to answer 2f without touching the algebra. If the table has not been memorised, setting the magnification to −1 and substituting takes about as long and is safer. There is a symmetry worth noticing behind the result: object and image distances are interchangeable in the lens formula, so the one place where a real image can equal its object in size is the place where the two distances are equal, and the formula then forces each of them to be 2f. A quick check on the other options is to ask what image each would give — infinity, a virtual one, and a diminished one respectively.

• A convex lens gives a real, inverted, same-size image only when the object is at twice the focal length.
• With f = 15 cm, that object distance is 30 cm, and the image forms 30 cm on the far side.
• The thin-lens formula is 1/v − 1/u = 1/f, and same-size-and-real corresponds to a magnification of −1.
• An object at the focus gives an image at infinity; an object inside the focus gives a virtual, erect, enlarged image.
• An object beyond twice the focal length gives a real, inverted, diminished image between f and 2f.

• Answering with the focal length itself instead of twice it.
• Forgetting that a real image needs the object outside the focal length.
• Dropping the minus sign in the magnification and treating a same-size real image as m = +1.

As a numerical fixing the object distance from a stated focal length and a described image, or in reverse — the position is given and the nature of the image is asked for.

• practice — not a real PYQ

  An object is placed 20 cm in front of a convex lens of focal length 20 cm. The image formed will be

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

  Answer(d) formed at infinity — the object sits exactly at the focus, so the emerging rays are parallel and meet only at infinity.

• practice — not a real PYQ

  A convex lens forms a real image twice the size of the object. If the focal length of the lens is 10 cm, the object distance is

  • (a)5 cm
  • (b)15 cm
  • (c)20 cm
  • (d)30 cm

  Answer(b) 15 cm — a real image twice the size means m = −2, so v = −2u; substituting in the lens formula gives u = −1·5f, that is 15 cm, which lies between f and 2f as the enlarged-real case requires.