If the magnification produced by a lens is +2, then the image is:
- (a)erect, virtual and smaller than the object.
- (b)inverted, real and smaller than the object.
- (c)erect, virtual and larger than the object.
- (d)inverted, real and larger than the object.
Correct — C, erect, virtual and larger than the object. Magnification carries two pieces of information at once, and the answer needs both to be read off. Its sign tells you the orientation: for a lens, a positive magnification means an erect image, and an erect image formed by a lens is always a virtual one, standing on the same side as the object with no light actually arriving there. Its size tells you the scale: a magnitude of 2 means the image is twice as tall as the object, so it is enlarged. Put the two together and +2 describes an erect, virtual, doubled image — the everyday behaviour of a magnifying glass. The geometry that produces exactly this is worth carrying: writing v = 2u in the thin-lens formula gives an object distance of half the focal length, with the image standing a full focal length away on the same side.
- (a)erect, virtual and smaller than the object. — Reads the plus sign correctly but ignores the number. A magnification of +2 means twice the height; the erect, virtual and smaller case is what a magnification such as +0.5 describes, which is the everyday behaviour of a concave lens.
- (b)inverted, real and smaller than the object. — Gets both halves wrong at once. Inverted and real go with a negative magnification, and diminished needs a magnitude below 1 — this option describes something like -0.5, not +2.
- (d)inverted, real and larger than the object. — The most tempting wrong answer, because the size is right. It quietly turns the plus into a minus. A real, inverted and doubled image is m = -2, which a convex lens gives when the object sits between the focus and twice the focal length.
The magnification of a thin lens is the ratio of image height to object height, and it equals the ratio of image distance to object distance, m = h'/h = v/u, with distances measured from the optical centre. Two features of the number matter. The sign fixes orientation and nature: positive means erect, and for a lens an erect image is virtual, while negative means inverted and real. The magnitude fixes size: greater than 1 is enlarged, less than 1 is diminished, exactly 1 is same-size. A concave lens acting on a real object can only ever return a positive magnification smaller than 1, so any magnitude above 1 has to come from a convex lens.
The commonest way to lose this mark is to treat the sign as decoration and answer only on the number 2, which lands on option (d). The habit that fixes it is to split the value in two before looking at the choices — read the sign, then read the size, then find the option that satisfies both. The stem's value is also a specific arrangement rather than a vague one. Setting v = 2u in 1/v - 1/u = 1/f gives an object distance of f/2 and an image distance of f on the object's own side, which is a magnifying glass held closer to the page than its focal length. The bank records an image flag against this question, but the stem as printed is self-contained and needs no figure.
- For a thin lens, magnification m = h'/h = v/u, with all distances measured from the optical centre.
- A positive magnification means an erect image, and an erect image formed by a lens is virtual; a negative magnification means an inverted, real image.
- A magnitude above 1 is an enlarged image, below 1 a diminished one, and exactly 1 a same-size image.
- A magnification of exactly +2 from a convex lens puts the object at half the focal length and the image one focal length away on the same side.
- A concave lens acting on a real object always gives a positive magnification less than 1 — erect, virtual and diminished, every time.
Sign first, size second. Answering on the size alone is what lands candidates on the inverted-and-real option.
- Answering on the magnitude alone and choosing the inverted, real, enlarged option.
- Assuming any enlarged image must be real, when an erect enlarged image is always virtual.
- Applying the mirror formula m = -v/u to a lens.
Either as this sign-and-size reading, or the other way round — the nature of the image is described and the sign of the magnification is asked for.
What is the magnification produced by a concave lens of focal length 10 cm, when an image is formed at a distance of 5 cm from the lens?
- (a) 2.0
- (b) 1.0
- (c) 0.5
- (d) 0.33
Answer(c) 0.5
The same quantity computed rather than interpreted, from the same exam season. A concave lens puts the object 10 cm away for an image at 5 cm, so v/u works out to 0·5 — positive and below 1, which is the only kind of magnification a concave lens can give.
CDS_GK_2021_II_Q12021Where should an object be placed in front of a convex lens to get a real and enlarged image of the object ?
- (a) At twice the focal length
- (b) At infinity
- (c) Between the principal focus and twice the focal length
- (d) Beyond twice the focal length
Answer(c) Between the principal focus and twice the focal length
The same table entered from the other end. That item asks where the object goes to make the image real and enlarged, which is the negative-magnification zone; this one hands you a positive magnification and asks what kind of image it describes.
- practice — not a real PYQ
The magnification produced by a lens is -0·5. The image formed is
- (a)erect, virtual and half the size of the object
- (b)inverted, real and half the size of the object
- (c)erect, virtual and twice the size of the object
- (d)inverted, real and twice the size of the object
Answer(b) inverted, real and half the size of the object — the minus sign gives inverted and real, and a magnitude of 0·5 halves the height.
- practice — not a real PYQ
For any real object placed in front of a concave lens, the magnification produced is
- (a)always negative and greater than 1 in magnitude
- (b)always positive and less than 1
- (c)always exactly +1
- (d)positive or negative depending on the object distance
Answer(b) always positive and less than 1 — a concave lens gives an erect, virtual, diminished image wherever the real object is placed.