Formula
Thin lens formula
1/v - 1/u = 1/f. Same sign convention as for mirrors. Power P = 1/f (in dioptres if f in metres).
-- NCERT Class 12 Physics, Ch. 9, p. 234Thin Lens Formula
8 MCQs1 revision card9-step worked example
Source: NCERT OpticsOfficial key: NTA-verifiedLast updated: 27 Sep 2026
Thin Lens Formula, explained for NEET
The thin lens formula is 1/v − 1/u = 1/f. The minus sign is the whole lesson. Mirror work leaves a groove — 1/v + 1/u = 1/f — and under pressure the hand writes a plus. Sign-convention inversion of u, v or f is a documented error for exactly this formula, and it is silent: you get a clean number that sits in the options list.
Fix the convention before you touch arithmetic. Optical centre at the origin. Light travels left to right, and that direction is positive. So a real object on the left has u negative — always, without exception, for a real object. A converging lens has f positive; a diverging lens has f negative. Choose this convention and hold it for the entire question.
NCERT Class 12 Physics Part 2, Chapter 9, page 234 states the relation in this signed form. Write u = −0.30 m, not u = 0.30 m, in the substitution line itself. Substituting a bare magnitude and hoping to reason about the sign afterwards is where the mark goes.
Read the answer the convention gives you. A positive v means the image is on the outgoing side — real, and for a single lens, inverted. A negative v means the image is on the object side — virtual and erect. That sign is physics, not bookkeeping: a question asking "real or virtual?" is answered by the sign of v and nothing else.
Two thin-lens facts worth holding. An object at 2f from a converging lens puts the image at 2f on the far side, same size. An object inside f gives a virtual image — the magnifying-glass case — and v comes out negative to tell you so.
Watch out for the rearrangement. From 1/v = 1/f + 1/u, the right-hand side is a sum of reciprocals. Invert it to get v. Reporting 1/v as v gives an answer off by orders of magnitude, and it will be on the option list.
Can you answer these Thin Lens Formula MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The thin lens formula relating object distance u, image distance v and focal length f is
Show answer and why every option is right or wrong
Answer: C. C is correct. The thin lens formula is 1/v − 1/u = 1/f, stated in this signed form in NCERT Class 12 Physics Part 2, Chapter 9, page 234.
Why A is wrong: A is wrong because it is the mirror formula, not the lens formula. The plus sign belongs to spherical mirrors; carrying it into lens work is the single most common sign-convention slip on this topic.
Why B is wrong: B is wrong because it reverses which reciprocal is subtracted. It gives −1/f, so every focal length comes out with the wrong sign.
Why D is wrong: D is wrong because the relation is additive in reciprocals, not multiplicative. No form of the lens formula involves a product of 1/v and 1/u.
MCQ 2Easy RecallPractice
Under the Cartesian sign convention used with the thin lens formula, distances along the principal axis are measured from the
Show answer and why every option is right or wrong
Answer: D. D is correct. For a thin lens all axial distances — u, v and f — are measured from the optical centre, which is taken as the origin. NCERT Class 12 Physics Part 2, Chapter 9, page 234.
Why A is wrong: A is wrong because the focal point is a location the formula solves for or uses as f, not the origin distances are measured from.
Why B is wrong: B is wrong because the centre of curvature belongs to the lens maker's description of a surface, not to the axial origin used in the lens formula.
Why C is wrong: C is wrong because a thin lens is treated as having negligible thickness, so its two surfaces and its optical centre are taken as one point. Measuring from a surface implies a thickness the thin-lens approximation discards.
MCQ 3Easy RecallPractice
For a real object placed in front of a thin lens, with light travelling in the positive direction, the object distance u is
Show answer and why every option is right or wrong
Answer: B. B is correct. A real object lies on the incoming side of the optical centre, against the direction taken as positive, so u is negative for every real object regardless of lens type. NCERT Class 12 Physics Part 2, Chapter 9, page 234.
Why A is wrong: A is wrong because it substitutes the bare magnitude. Writing u as positive for a real object is the sign inversion that this formula punishes most often; it flips the sign of the computed f or v.
Why C is wrong: C is wrong because the sign of u depends only on which side of the optical centre the object sits, not on the lens being converging or diverging. The lens type sets the sign of f, not of u.
Why D is wrong: D is wrong because the nature of the image is an output of the calculation. The sign of u is fixed by the object's position before any arithmetic happens.
MCQ 4Direct ApplicationPractice
Applying the thin lens formula to a problem yields a negative value of v. This tells you that the image is
Show answer and why every option is right or wrong
Answer: A. A is correct. A negative v places the image on the incoming side of the optical centre — the object's side — and an image there is not formed by actual converging rays, so it is virtual. NCERT Class 12 Physics Part 2, Chapter 9, page 234.
Why B is wrong: B is wrong because it pairs the correct side with the wrong nature. On the object's side of a single thin lens, rays are diverging, so the image cannot be real.
Why C is wrong: C is wrong because a real image on the far side corresponds to positive v. This option reads the sign backwards.
Why D is wrong: D is wrong because the far side is the positive-v side. An image cannot be both at negative v and on the far side; the two descriptions contradict each other.
MCQ 5Direct ApplicationPractice
A real object is placed 3.00 × 10⁻¹ m from a thin converging lens of focal length 2.00 × 10⁻¹ m. The image distance v is
Show answer and why every option is right or wrong
Answer: D. D is correct. With u = −3.00 × 10⁻¹ m and f = +2.00 × 10⁻¹ m, 1/v = 1/f + 1/u = 5.00 − 3.33 = 1.67 m⁻¹, so v = +6.00 × 10⁻¹ m — positive, meaning the far side. NCERT Class 12 Physics Part 2, Chapter 9, page 234.
Why A is wrong: A is wrong because 1.67 m⁻¹ is the value of 1/v, not v. Reporting the reciprocal sum as the answer without inverting leaves the result in the wrong units and off by orders of magnitude.
Why B is wrong: B is wrong because it comes from using the mirror form 1/v + 1/u = 1/f with a signed u, giving 1/v = 5.00 + 3.33 = 8.33 m⁻¹. Carrying the mirror's plus sign into lens work is the classic sign-convention inversion on this formula.
Why C is wrong: C is wrong because it takes the right magnitude but reports the image on the object's side. The computed v is positive, which places the image on the outgoing side.
MCQ 6Direct ApplicationPractice
A real object is placed 1.00 × 10⁻¹ m from a thin diverging lens of focal length of magnitude 2.00 × 10⁻¹ m. The image distance v is
Show answer and why every option is right or wrong
Answer: B. B is correct. A diverging lens has f = −2.00 × 10⁻¹ m, and u = −1.00 × 10⁻¹ m, so 1/v = 1/f + 1/u = −5.00 − 10.0 = −15.0 m⁻¹, giving v = −6.67 × 10⁻² m. NCERT Class 12 Physics Part 2, Chapter 9, page 234.
Why A is wrong: A is wrong because it substitutes both f and u as positive magnitudes: 1/v = 5.00 + 10.0 = 15.0 m⁻¹ gives +6.67 × 10⁻² m, the right size with the wrong sign. Substituting magnitudes rather than signed values is what produces it.
Why C is wrong: C is wrong because it takes f as positive for a diverging lens: 1/v = 5.00 − 10.0 = −5.00 m⁻¹ gives v = −2.00 × 10⁻¹ m. A diverging lens must carry f < 0 throughout.
Why D is wrong: D is wrong because it uses the correct negative f but leaves u positive: 1/v = −5.00 + 10.0 = +5.00 m⁻¹ gives v = +2.00 × 10⁻¹ m. Both u and f must be signed in the substitution line.
MCQ 7Concept TrapPractice
A student applies the thin lens formula to a converging lens with a real object and obtains v with the same sign as u. Without repeating the arithmetic, what does that result indicate about the object's position?
Show answer and why every option is right or wrong
Answer: A. A is correct. For a real object u is negative, so v sharing that sign means v is negative — a virtual image, which a converging lens produces only when the object lies inside the focal length. NCERT Class 12 Physics Part 2, Chapter 9, page 234.
Why B is wrong: B is wrong because an object at 2f gives a real image at 2f on the far side, so v is positive while u is negative.
Why C is wrong: C is wrong because an object between f and 2f still gives a real image beyond 2f on the far side, with v positive.
Why D is wrong: D is wrong because an object beyond 2f gives a real image with positive v, opposite in sign to u.
MCQ 8CalculationPractice
A thin lens forms an image of a real object on a screen placed 4.00 × 10⁻¹ m beyond the lens, with the object 4.00 × 10⁻¹ m in front of it. The lens is then replaced by one of half the focal length, the object left where it is. The new image distance is
Show answer and why every option is right or wrong
Answer: C. C is correct. The first arrangement gives 1/f = 1/(+4.00 × 10⁻¹) − 1/(−4.00 × 10⁻¹) = 5.00 m⁻¹, so f = 2.00 × 10⁻¹ m; the new lens has f = 1.00 × 10⁻¹ m, and 1/v = 10.0 − 2.50 = 7.50 m⁻¹, giving v = 1/7.50 = 1.33 × 10⁻¹ m. NCERT Class 12 Physics Part 2, Chapter 9, page 234.
Why A is wrong: A is wrong because it assumes the image stays put when the focal length changes. An image on a screen at the object distance is the 2f–2f case for the original lens only; halving f moves the image.
Why B is wrong: B is wrong because it halves the image distance along with the focal length. The lens formula is not linear in f, so v does not scale with f.
Why D is wrong: D is wrong because 1.00 m does not follow from the lens formula with the new focal length. The first image gives 1/f = 1/v − 1/u = 2.50 + 2.50 = 5.00 m⁻¹, so f = 0.200 m; the new lens has f = 0.100 m, and 1/v = 1/f + 1/u = 10.0 − 2.50 = 7.50 m⁻¹, so v = 0.133 m.
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Thin Lens Formula: quick recall before you leave
How do you solve a Thin Lens Formula question? A worked example
- 1
Given
A thin converging lens of focal length 1.50 × 10⁻¹ m. A real object on the principal axis, 1.00 × 10⁻¹ m from the optical centre. (Both distances are measurements, written in scientific notation so the trailing zeros are unambiguous — three significant figures each.)
- 2
Required
The image distance v, and whether the image is real or virtual.
- 3
Concept
The object sits inside the focal length of a converging lens. The refracted rays still diverge after the lens, so they cannot meet; the image is located by projecting them backwards, which puts it on the object's side. The formula will report this as a negative v — we do not need to decide it in advance, only to read the sign honestly at the end.
- 4
Formula
1/v − 1/u = 1/f, rearranged to1/v = 1/f + 1/u. - 5
Substitution
Sign the values before they enter the line. Light travels in the positive direction; the real object lies against it, so
u = −1.00 × 10⁻¹ m. The lens converges, sof = +1.50 × 10⁻¹ m.1/v = 1/(+1.50 × 10⁻¹ m) + 1/(−1.00 × 10⁻¹ m) - 6
Calculation
1/v = 6.67 m⁻¹ − 10.0 m⁻¹ = −3.33 m⁻¹
Now invert — this is the step that gets skipped:v = 1/(−3.33 m⁻¹) = −3.00 × 10⁻¹ m
No exact constants enter this calculation; every quantity used is a measured value carrying three significant figures, so the answer is reported to three. - 7
Final answer
v = −3.00 × 10⁻¹ m. The negative sign places the image 3.00 × 10⁻¹ m from the lens on the same side as the object, so the image is virtual. (A magnifying glass held closer to the print than its focal length: this is that arrangement.) - 8
Common trap
Two sign failures live in step 5. The first is writing
u = +1.00 × 10⁻¹ mbecause the object is "1.00 × 10⁻¹ m away" — a magnitude substituted where a signed value belongs. That gives1/v = 6.67 + 10.0 = 16.7 m⁻¹andv = +6.00 × 10⁻² m, a real image that does not exist. The second is importing the mirror formula's plus sign and writing1/v = 1/f − 1/u, which produces the same wrong number by a different route. Both leave a clean, plausible figure. Sign the values in the substitution line itself; do not reason about the sign afterwards. - 9
Similar NEET-style question
A thin diverging lens of focal length of magnitude 1.20 × 10⁻¹ m has a real object placed 2.40 × 10⁻¹ m from it. Find the image distance and state whether the image is real or virtual. (Watch the sign of f before you start: the magnitude is given, the sign is not.)
What to remember before solving Thin Lens Formula questions
Which Thin Lens Formula formulas do you need for NEET?
1 formula — click to collapse
Thin lens formula
Thin lens formula. Power P = 1/f (in dioptres if f in metres).
| Symbol | Quantity | SI Unit |
|---|---|---|
| v | image distance | m |
| u | object distance | m |
| f | focal length | m |
Valid when
- Thin lens
- Paraxial rays
- Sign convention used
Where do students lose marks on Thin Lens Formula?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
2 items — click to collapse
Category: Sign Convention
Student inverts the sign of u, v, or f in mirror/lens formulas. Convention: distances measured from pole/optical centre; positive in direction of light propagation.
When it triggers
Mirror or lens problem requiring formula application.
How to avoid
Cartesian sign convention: pole at origin; light travels in +x direction; distances measured along axis are signed. Concave mirror f<0 on convention where light incident from left (or vice versa per text). Apply ONE convention throughout.
Root cause: sign error
Correction
Use Cartesian convention consistently: pole/optical-centre at origin, light travels in +x direction. Object usually has u < 0 (left of pole). Concave mirror f < 0 (most conventions). Apply ONE convention throughout.
More in Optics: 7 exam traps and mistakes · 10 formulas · 7 question patterns from its other lessons.
Thin Lens Formula questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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