Optics is a reliable scoring zone in the CDS & OTA General Science paper, and lenses sit right at its heart. Every year the examiner asks about image formation, the lens formula, magnification and the power of a lens in dioptres. This page rebuilds the topic from first principles so you can predict an image, plug into the formula and pick the right option fast.
Why lenses are a dependable CDS scorer
In the CDS General Science (Part B) section, physics regularly contributes a cluster of questions, and optics is one of its most predictable themes. Within optics, lenses are a favourite because a single idea — how a transparent curved surface bends light — generates both conceptual questions (which lens corrects which eye defect?) and quick numericals (find the power, find the image distance).
A lens is a piece of transparent material, usually glass, bounded by two surfaces of which at least one is curved. It works by refraction: light changes speed and direction as it passes from air into glass and back out. Because the two faces are curved, different rays bend by different amounts and converge to, or appear to diverge from, a single point.
A lens does not create light; it only redirects existing rays by refraction. Mirrors do the same job by reflection. Confusing the two is the most common reason candidates lose easy optics marks.
The payoff for mastering lenses is that the rules are tight and repeatable. Once you fix the sign convention and two formulas in your head, most CDS lens questions collapse into a few seconds of substitution. That speed is valuable: the science section is wide, so banking quick, certain marks in optics frees up time for the longer biology and chemistry recall items.
Convex and concave: the two lens families
Every lens you meet in CDS is one of two basic types, defined by whether it is thicker or thinner at the middle.
Convex (converging) lens
A convex lens is thicker at the centre than at the edges. Parallel rays striking it are bent inwards and meet at a real point called the principal focus. Because it brings rays together, it is also called a converging lens. A magnifying glass is a convex lens.
Concave (diverging) lens
A concave lens is thinner at the centre and thicker at the edges. Parallel rays spread out after passing through it, appearing to come from a focus on the same side as the incoming light. It is therefore a diverging lens and always forms a small, erect, virtual image.
Convex = converging = thick middle = real focus on the far side. Concave = diverging = thin middle = virtual focus on the near side. The optical centre is the point through which a ray passes undeviated.
Two terms recur in every question. The principal axis is the straight line through the centres of curvature of the two faces. The focal length (f) is the distance from the optical centre to the principal focus. A short focal length means the lens bends light strongly; a long focal length means it bends light gently.
Image formation by a convex lens
The nature of the image from a convex lens depends entirely on where the object sits relative to the focus (F) and twice the focal length (2F). Learn this table cold — CDS loves to test it directly.
- Object at infinity → image at F, real, inverted, highly diminished (a point).
- Object beyond 2F → image between F and 2F, real, inverted, diminished.
- Object at 2F → image at 2F, real, inverted, same size.
- Object between F and 2F → image beyond 2F, real, inverted, magnified.
- Object at F → image at infinity, real, inverted, highly magnified.
- Object between F and lens → image on the same side, virtual, erect, magnified.
A convex lens gives a virtual, erect, enlarged image only when the object is closer than the focus — that is exactly the magnifying-glass setting. In every other position the image is real and inverted.
A concave lens, by contrast, is far simpler: no matter where the object is, the image is always virtual, erect and diminished, formed between the focus and the lens on the same side as the object. If a question describes a lens that always shrinks the image, it must be concave.
The Cartesian sign convention
Numericals only work if you apply signs consistently. CDS follows the standard Cartesian sign convention, measured from the optical centre along the principal axis.
- All distances are measured from the optical centre.
- Distances measured in the direction of incident light (to the right) are positive; those measured against it (to the left) are negative.
- Heights above the principal axis are positive; heights below are negative.
For a real object the object distance u is always negative. A convex lens has positive f; a concave lens has negative f. A positive image distance v means a real image on the far side; a negative v means a virtual image on the same side as the object.
Forgetting that u is negative is the single biggest error in lens numericals. Candidates write u = 20 cm instead of u = −20 cm and get a wrong, often impossible, answer. Always insert the sign before you substitute.
Keep the convention identical for every problem. The examiner sometimes offers a distractor option that matches the answer you would get from a sign slip, so a disciplined convention protects your marks.
Lens formula and magnification
Two relations do almost all the numerical work in this topic.
The lens formula
The lens formula connects the object distance (u), the image distance (v) and the focal length (f):
1/v − 1/u = 1/f
Here v = image distance, u = object distance, f = focal length, all with their proper signs.
Magnification
The magnification (m) tells you how big the image is compared with the object, and whether it is erect or inverted:
m = height of image / height of object = v / u
A positive m means a virtual, erect image; a negative m means a real, inverted image. |m| > 1 is magnified, |m| < 1 is diminished.
Note the difference from mirrors, where m = −v/u. For lenses the magnification is simply v/u with no extra minus sign. Mixing up the two formulas is a frequent trap, so anchor the lens version firmly: lens → m = v/u.
Power of a lens and the dioptre
The power of a lens measures its ability to bend (converge or diverge) light. A lens with a short focal length bends light strongly and so has a large power.
P = 1 / f (with f in metres)
The SI unit of power is the dioptre (D), where 1 D is the power of a lens of focal length 1 metre. A convex lens has positive power; a concave lens has negative power.
Because f goes in metres, a lens of focal length 20 cm = 0.2 m has power P = 1/0.2 = +5 D if convex. Always convert centimetres to metres first.
Lenses in contact
When two thin lenses are placed in contact, their powers simply add:
P = P1 + P2 and equivalently 1/f = 1/f1 + 1/f2
If a convex lens (+4 D) and a concave lens (−6 D) are combined, the net power is +4 + (−6) = −2 D — the combination behaves as a diverging lens. Just add the powers with their signs; never add focal lengths directly.
Lenses in the human eye and its defects
CDS frequently links lenses to vision defects, because the eye contains a natural convex lens whose focal length the ciliary muscles adjust — a process called accommodation.
Myopia (short-sightedness)
A myopic eye sees near objects clearly but distant objects appear blurred, because the image forms in front of the retina. It is corrected by a concave (diverging) lens of suitable negative power, which pushes the image back onto the retina.
Hypermetropia (long-sightedness)
A hypermetropic eye sees distant objects clearly but near objects appear blurred, because the image forms behind the retina. It is corrected by a convex (converging) lens of positive power.
Myopia → concave lens (− power). Hypermetropia → convex lens (+ power). A simple memory hook: myopia needs the lens that is open (thin) at the centre — the concave lens.
A third defect, presbyopia, is the age-related loss of accommodation and often needs bifocal lenses combining concave and convex parts. These eye-defect links are pure recall, so memorise the lens-defect pairing and you secure those marks instantly.
Worked example: finding image distance and power
Let us run a complete numerical with the sign convention applied carefully.
An object is placed 30 cm in front of a convex lens of focal length 20 cm. Find the position and nature of the image, the magnification, and the power of the lens.
The image is real, inverted and twice the object size, sitting 60 cm beyond the lens, and the lens has a power of +5 dioptres. This matches the rule that an object between F and 2F (here 30 cm, between 20 and 40 cm) gives a magnified real image beyond 2F.
Common mistakes that cost marks
Lens questions are easy once you avoid a short list of recurring slips. Scan this before the exam.
- Sign errors: writing u as positive, or giving a concave lens a positive focal length. Fix the signs first, always.
- Wrong magnification formula: using m = −v/u (the mirror rule) instead of m = v/u for a lens.
- Forgetting unit conversion: computing power with f in centimetres. P needs f in metres.
- Confusing the two lenses: assigning a concave lens to hypermetropia, or assuming a concave lens can magnify. It cannot — it always diminishes.
Adding focal lengths directly when two lenses are in contact. You must add powers (or reciprocals of focal lengths), not the focal lengths themselves.
A final discipline: once you compute an image, sanity-check it against the qualitative table. If your formula gives a real, magnified image but the object sat closer than the focus, you have made a sign slip somewhere — the physics and the arithmetic must agree.
Previous-year style question, solved
Here is a question in the exact style CDS uses for lens power, worked through fully.
Q. The power of a lens is −2.5 D. What is the focal length and nature of the lens?
Answer: Using P = 1/f, we get f = 1/P = 1/(−2.5) = −0.4 m = −40 cm. The negative sign shows the lens is concave (diverging), with a focal length of 40 cm.
Whenever a question gives the power directly, read the sign first: a negative power means a concave lens, a positive power means a convex lens. Then convert the focal length to centimetres if the options demand it.
Many CDS items stop at identifying the lens type from the sign of the power, so even if the arithmetic feels rushed, the sign alone often points you to the correct option.
Quick revision before the exam
Run through these final checks the night before, and the lens cluster becomes near-automatic.
- Convex = converging, thick middle, positive f and positive power; concave = diverging, thin middle, negative f and negative power.
- Lens formula: 1/v − 1/u = 1/f, with the Cartesian sign convention (real object u is negative).
- Magnification m = v/u; positive → virtual & erect, negative → real & inverted.
- Power P = 1/f (f in metres), unit dioptre; lenses in contact add powers: P = P1 + P2.
- Myopia → concave lens; hypermetropia → convex lens.
- A convex lens magnifies only when the object is inside the focus; a concave lens always diminishes.
Memorise the image-position table, lock the two formulas with their signs, and practise five or six numericals until substitution is reflexive. With that, the lens questions in the CDS science paper turn into some of the quickest, surest marks on the whole sheet.
Frequently asked questions
What is the difference between a convex and a concave lens?
A convex lens is thicker at the centre and converges parallel light to a real focus, so it can magnify. A concave lens is thinner at the centre, diverges light to a virtual focus, and always forms a diminished, erect image.
What is the lens formula used in CDS numericals?
The lens formula is 1/v − 1/u = 1/f, where v is the image distance, u the object distance and f the focal length, all taken with the Cartesian sign convention. For a real object u is negative.
How is the power of a lens calculated and what is its unit?
Power P = 1/f with the focal length f in metres. Its SI unit is the dioptre (D). A convex lens has positive power and a concave lens has negative power.
Which lens corrects myopia and which corrects hypermetropia?
Myopia (short-sightedness) is corrected by a concave (diverging) lens of negative power, while hypermetropia (long-sightedness) is corrected by a convex (converging) lens of positive power.
How do you find the combined power of two lenses in contact?
Add their individual powers with signs: P = P1 + P2. For example, a +4 D convex lens and a −6 D concave lens in contact give a net power of −2 D, behaving as a diverging combination.
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