EPFO APFC 2023 General Studies · Question 51 of 120

Correct — B, (b) act as a concave lens. When two thin lenses are placed in contact, their powers simply add, so the combination behaves as a single lens whose power is the algebraic sum of the two. Here that sum is +2·0 D + (–2·5 D) = –0·5 D. The result is negative, and a negative power is the definition of a diverging lens — that is, a concave one. So the combination acts as a concave lens. Power is the reciprocal of the focal length in metres, P = 1/f, and it is measured in dioptres. A lens of one dioptre has a focal length of one metre. The two lenses in this question therefore have focal lengths of 1/2·0 = +0·5 m, or +50 cm, for the converging one, and 1/(–2·5) = –0·4 m, or –40 cm, for the diverging one. The combination has a focal length of 1/(–0·5) = –2 m, a diverging lens of focal length two metres, which is a weak one — but weak and diverging is still diverging. The physical reading is worth more than the arithmetic. Power measures how strongly a lens bends light: the shorter the focal length, the larger the power. The concave lens here has the shorter focal length in magnitude, 40 cm against 50 cm, so it is the stronger of the two and it decides the character of the pair. The convex lens does not cancel it; it only weakens it, leaving a net divergence of half a dioptre. Reverse the numbers — a +2·5 D lens with a –2·0 D lens — and the same reasoning gives +0·5 D and a converging combination. The rule to carry away is that the lens with the greater magnitude of power wins, and equal magnitudes of opposite sign give a combination of zero power, which is a plane plate that bends nothing. Note the printing: the booklet sets the decimal point as a raised middle dot, so the powers appear as +2·0 D and –2·5 D, and the minus sign is printed at en-dash length. Both are reproduced here as printed, and neither is an error.

• (a)act as a convex lens — This is what the combination would do if the net power came out positive, and it is the answer a candidate reaches by adding the magnitudes instead of the signed values, or by assuming that the convex lens named first in the stem sets the character of the pair. The sign is the whole of the physics: +2·0 D and –2·5 D combine to –0·5 D, not to +4·5 D or +0·5 D. It is worth being able to state the boundary case as well — if the second lens were exactly –2·0 D, the combination would have zero power and behave like a flat sheet of glass, deviating light not at all rather than converging it. Only when the positive power exceeds the negative in magnitude does the pair converge.
• (c)act as a simple mirror — A lens works by refraction — light passes through it and is bent at the two surfaces — while a mirror works by reflection, sending light back on the side it came from. No arrangement of two lenses in contact can turn one into the other, whatever their powers, because neither surface is silvered and nothing sends the light back. The formula for the combination confirms it in its own terms: the sum of two lens powers is a lens power, and it describes a lens of focal length 1/P, not a reflecting surface. The one situation in which a lens and a mirror are described by a single equivalent power is when a lens is actually silvered on one face, which is a quite different arrangement and is not what the stem describes.
• (d)not form any image — A diverging lens forms an image perfectly well; it simply forms a virtual one. Rays from a real object leave a concave lens diverging, so they never actually meet, but their backward extensions do, and the eye sees an image that is virtual, erect and diminished, lying on the same side of the lens as the object and always between the optical centre and the focus. For the combination here, with a focal length of two metres, an object placed one metre away would give an image about 67 centimetres from the lens on the object's side, upright and about a third smaller. The only case in which a combination genuinely forms no image at all is a net power of exactly zero, when the pair acts as a plane plate — and that would need the two powers to be equal in magnitude, which they are not.

The power of a lens is the reciprocal of its focal length expressed in metres, and it is measured in dioptres, so that a lens of focal length half a metre has a power of two dioptres. The sign convention carries the whole classification: a converging or convex lens has a positive focal length and a positive power, while a diverging or concave lens has a negative focal length and a negative power. Power is the more convenient quantity for combinations because, for thin lenses placed in contact, powers add algebraically — the equivalent power is the sum of the individual powers, which is the same statement as the reciprocal of the equivalent focal length being the sum of the reciprocals of the individual focal lengths. That additivity is why an optician writes a prescription in dioptres and why the strength of a compound lens can be worked out in one line. Three consequences follow and are worth holding together. First, the sign of the sum decides whether the combination converges or diverges, so a pair made of one lens of each kind behaves like whichever is stronger. Second, magnitude measures bending power, so a shorter focal length means a larger power, and a small focal length lens dominates a large one. Third, a combination of zero net power is not a lens that fails to work but a plate that does not deviate light at all. Achromatic doublets are built on exactly this arithmetic: a converging crown glass lens is cemented to a weaker diverging flint glass lens so that the pair still converges while the colour spreading of the two largely cancels.

The general science block of this paper asks for a physical conclusion rather than a numerical answer — the options here are descriptions, not values — which is a shape that rewards candidates who understand the sign convention and punishes those who have memorised a formula without it. The arithmetic takes five seconds; the decision about what a negative power means is the actual question. Optics recurs across the science block in this paper, with a mirror item close by, so the pair of sign conventions, one for lenses and one for mirrors, is worth revising together. Note the printed form of the item as well. The decimals are set with a raised middle dot rather than a full stop, as they are on four other questions in this booklet, the minus sign is printed at en-dash length, and the stem is an incomplete sentence that each option completes, with no question mark at the end. None of these is a misprint, and option (d) contains a negation of its own that has nothing to do with the ask.

• Power P = 1/f with f in metres, measured in dioptres; a convex or converging lens has positive power and a concave or diverging lens has negative power.
• For thin lenses in contact the powers add algebraically: P = P₁ + P₂, equivalently 1/F = 1/f₁ + 1/f₂.
• +2·0 D combined with –2·5 D gives –0·5 D, so the combination is diverging and behaves as a concave lens of focal length 1/(–0·5) = –2 m.
• The individual focal lengths here are +50 cm for the +2·0 D lens and –40 cm for the –2·5 D lens; the shorter magnitude means the concave lens is the stronger one and sets the character of the pair.
• Two lenses of equal and opposite power give a combination of zero power, which behaves like a plane sheet of glass rather than as a lens that fails to work.
• A concave lens always forms a virtual, erect and diminished image of a real object, located between the optical centre and the focus on the same side as the object.

• Adding the magnitudes of the two powers instead of their signed values; the whole answer is in the sign of the sum
• Assuming the first-named or the stronger-sounding lens sets the character of the pair, when it is the larger magnitude of power that decides
• Confusing power with focal length — a larger power means a shorter focal length, so the numerically smaller focal length is the stronger lens
• Reading a negative power as meaning no image; a diverging lens forms a virtual, erect, diminished image for every position of a real object
• Applying the simple sum of powers to lenses that are separated rather than in contact, where a third term involving the separation appears
• Using focal lengths in centimetres in the power formula; the focal length must be in metres for the answer to come out in dioptres

Optics in this paper is set as a short conceptual item — a combination of powers, or the image formed in a named configuration — with descriptive rather than numerical options. The examiner tests the sign convention and the standard image cases, not algebraic manipulation, so the highest-yield preparation is the table of object positions against image characteristics for a convex lens, a concave lens, a concave mirror and a convex mirror, together with the additivity of powers in dioptres.

No directly related past PYQ was found.

• practice — not a real PYQ

  A convex lens of power +5 D is placed in contact with a concave lens of power –3 D. The focal length of the combination is

  • (a)+50 cm
  • (b)+25 cm
  • (c)–50 cm
  • (d)–12·5 cm

  Answer(a) +50 cm — the powers add algebraically, giving +5 D + (–3 D) = +2 D, and the focal length is the reciprocal of the power in metres, so f = 1/2 = 0·5 m = +50 cm. The sign is positive, so the combination converges, the stronger convex lens having decided the character of the pair.

• practice — not a real PYQ

  A person can see distant objects clearly only when they are brought within 50 cm of the eye. The power of the corrective lens required is

  • (a)+2 D
  • (b)–2 D
  • (c)+0·5 D
  • (d)–0·5 D

  Answer(b) –2 D — the far point has moved in to 50 cm, which is myopia, and the correcting lens must form an image of a distant object at that far point. A diverging lens of focal length –50 cm does so, and its power is 1/(–0·5 m) = –2 D. A positive power would be prescribed for hypermetropia, where the near point has moved outwards.