NDA Physics · Light and Optics

Lenses and the Lens Formula

A lens refracts light through two curved surfaces: a convex (converging) lens bends rays together, a concave (diverging) lens spreads them apart. The lens formula 1/v − 1/u = 1/f, power P = 1/f in dioptres, the lens maker's equation, and magnification together fix every image.

Why this matters

Twelve PYQs, and the bank's cleanest source of numeric marks — power calculations and combinations appear almost every year. The recurring tests are: P = 1/f (with f in metres), adding powers for lenses in contact, the lens maker's equation, the New Cartesian sign convention, and the fixed fact that a concave lens only ever gives a virtual, erect, diminished image. All three HARD questions in the chapter involving lenses sit here.

Concept 1 of 6

Convex and concave lenses

Intuition

A convex lens is thick in the middle and squeezes parallel rays to a point — it converges light and can make real images. A concave lens is thin in the middle and fans rays apart — it diverges light and can only ever make a virtual, smaller, upright image. That single fact about the concave lens settles a surprising number of questions.

Definition

  • Convex (converging) lens: thicker at the centre; brings parallel rays to a real focus. It can form real or virtual images, and images larger, equal, or smaller than the object, depending on object position.
  • Concave (diverging) lens: thinner at the centre; spreads parallel rays so they appear to come from a virtual focus. It always forms a virtual, erect, diminished image — never a real one.

A convex lens can spread out a beam of sunlight (it refracts the rays toward a focus and beyond); a plane mirror cannot.

F2FF'2F'OobjectimageObject beyond 2F: real, inverted, diminished image between F′ and 2F′

A convex lens converges light. The parallel ray bends through the far focus; the ray through the optic centre goes straight. Where they cross is the image.

Worked example

State which statement is NOT correct: (i) a convex lens can produce both real and virtual images; (ii) a concave lens can produce both real and virtual images; (iii) a concave lens always produces a diminished image.
  1. A convex lens does produce real images (object beyond f) and virtual images (object inside f) — (i) is correct.
  2. A concave lens can NEVER produce a real image — it always gives a virtual one — so (ii) is the incorrect statement.
  3. A concave lens always gives a diminished image — (iii) is correct.
Answer:Statement (ii) is NOT correct — a concave lens produces only virtual images.
Practice this conceptself-check · 4 quick reps

Try it yourself

Which device can spread the light energy from the Sun over a larger area: a plane mirror, a convex lens, or a concave lens?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    A converging lens is convex or concave?
  2. 2.
    Can a concave lens ever form a real image?
  3. 3.
    Nature of every concave-lens image?
  4. 4.
    Which lens is thicker at the centre?

From the bank · past-year question

Example 1Light and OpticsMODERATE
Which one of the following statements regarding lenses is not\textbf{\text{not}} correct?

[Q71 · Sep · 2019]

A concave lens has no real-image setting

Unlike a concave MIRROR (which gives real images for most positions), a concave LENS gives a virtual, erect, diminished image for EVERY object position. Any claim that a concave lens makes real images is the wrong statement.

Concept 2 of 6

Lens formula and the sign convention

Intuition

The lens formula links object distance, image distance and focal length, just like the mirror formula but with a minus sign in a different place. The New Cartesian convention does the bookkeeping: measure everything from the optic centre, take the incident-light direction as positive, and a convex lens gets a positive focal length while a concave lens gets a negative one.

Definition

Lens formula: 1v1u=1f\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} (New Cartesian sign convention; distances measured from the optic centre).

  • Convex lens: ff is positive. Concave lens: ff is negative.
  • The same general FORM 1/v±1/u=1/f1/v \pm 1/u = 1/f underlies both mirror and lens — the convention differs because light is transmitted (lens) versus reflected (mirror), so the formula 'applies to spherical mirrors as well as spherical lenses'.

Lens formula

1v1u=1f\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}
  • uobject distance from optic centre (−ve for a real object)
  • vimage distance from optic centre
  • ffocal length (+ve convex, −ve concave)

Worked example

An object is placed 30 cm in front of a convex lens of focal length 20 cm. Where is the image formed?
  1. Sign convention: u=30u = -30 cm, f=+20f = +20 cm.
  2. 1v=1f+1u=120+130=360260=160\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{20} + \dfrac{1}{-30} = \dfrac{3}{60} - \dfrac{2}{60} = \dfrac{1}{60}.
  3. So v=+60v = +60 cm — a real image 60 cm beyond the lens.
Answer:v=+60v = +60 cm (real, on the far side).
Practice this conceptself-check · 4 quick reps

Try it yourself

Under the New Cartesian sign convention, what sign do the focal lengths of a convex lens and a concave lens carry?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Sign of f for a convex lens?
  2. 2.
    Sign of f for a concave lens?
  3. 3.
    In the lens formula, a real object has what sign of u?
  4. 4.
    Lens formula relating u, v, f?

From the bank · past-year question

Example 2Light and OpticsMODERATE
According to the New Cartesian Sign Convention, which one of the following is correct in respect of the formula 1f=1v+1u\frac{1}{f} = \frac{1}{v} + \frac{1}{u}, where symbols have their usual meanings ?

[Q68 · Apr · 2021]

Lens uses 1/v − 1/u; mirror uses 1/v + 1/u

The two formulae look almost identical — a sign on the 1/u term is the only difference. Mixing them up flips your image distance. Memorise: lens is MINUS, mirror is PLUS.

Concept 3 of 6

Power of a lens — the dioptre

Intuition

Power tells you how strongly a lens bends light: a short focal length means a powerful lens. Power is just one over the focal length, with the focal length in metres — and it is measured in dioptres. A convex lens has positive power; a concave lens has negative power.

Definition

Power P=1fP = \dfrac{1}{f} with ff in metres; unit: dioptre (D).

  • Convex lens: P>0P > 0. Concave lens: P<0P < 0.
  • A short focal length ⟹ large power (strongly bending).
  • Watch the units: f given in cm must be converted to metres first (fcm=f/100mf\,\text{cm} = f/100\,\text{m}).

Power of a lens

P=1f(in metres)P = \dfrac{1}{f\,(\text{in metres})}
  • Ppower (dioptre, D)
  • ffocal length in METRES

Worked example

A convex lens has a focal length of 25 cm. What is its power?
  1. Convert: f=25f = 25 cm =0.25= 0.25 m.
  2. P=1/f=1/0.25=+4P = 1/f = 1/0.25 = +4 D.
  3. Positive because the lens is convex.
Answer:+4 dioptre.
Practice this conceptself-check · 4 quick reps

Try it yourself

The focal length of a concave lens is 0.5 m. What is its power?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Power of a lens with f = 50 cm?
  2. 2.
    Unit of lens power?
  3. 3.
    Sign of power for a concave lens?
  4. 4.
    Power of a lens with f = 20 cm?

From the bank · past-year question

Example 3Light and OpticsEASY
Power of a lens of focal length 25 cm is

[Q132 · Sep · 2021]

Convert cm to metres before computing power

P = 1/f needs f in METRES. f = 25 cm gives P = 1/0.25 = 4 D, not 1/25 = 0.04 D. Forgetting the conversion is the standard power-question trap.

Concept 4 of 6

Lens maker's equation

Intuition

How do you build a lens of a chosen focal length? The lens maker's equation says it depends on two things: how much the glass slows light (its refractive index) and how sharply the two faces are curved (their radii). Make the surfaces more curved or the glass more refractive, and you get a more powerful lens.

Definition

Lens maker's equation: 1f=(n1)(1R11R2)\dfrac{1}{f} = (n - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right), for a thin lens in air.

  • nn is the refractive index of the lens material; R1,R2R_1, R_2 are the radii of curvature of the two faces (signed by convention).
  • For a double convex lens, R1>0R_1 > 0 and R2<0R_2 < 0, so the two terms add.
  • Higher nn or smaller radii ⟹ shorter focal length ⟹ greater power.

Lens maker's equation

1f=(n1)(1R11R2)\dfrac{1}{f} = (n - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)
  • nrefractive index of the lens material
  • R_1, R_2radii of curvature of the two faces (signed)
  • ffocal length

Worked example

A double convex lens has faces of radii 20 cm and 20 cm, made of glass with n = 1.5. Find its power.
  1. Double convex: R1=+20R_1 = +20 cm, R2=20R_2 = -20 cm.
  2. 1f=(1.51)(120120)=0.5×220=0.5×110=120cm1\dfrac{1}{f} = (1.5 - 1)\left(\dfrac{1}{20} - \dfrac{1}{-20}\right) = 0.5 \times \dfrac{2}{20} = 0.5 \times \dfrac{1}{10} = \dfrac{1}{20}\,\text{cm}^{-1}.
  3. So f=20f = 20 cm =0.2= 0.2 m, and P=1/0.2=+5P = 1/0.2 = +5 D.
Answer:f = 20 cm; power = +5 D.
Practice this conceptself-check · 4 quick reps

Try it yourself

The faces of a double convex lens have radii of curvature 10 cm and 20 cm, with glass of refractive index 1.5. Find the power of the lens.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    What two material/shape factors set a lens's focal length?
  2. 2.
    Does a higher refractive index give a longer or shorter focal length?
  3. 3.
    For a double convex lens, R₂ carries which sign?
  4. 4.
    Lens maker's equation: 1/f = ?

From the bank · past-year question

Example 4Light and OpticsHARD
The radii of curvature of the faces of a double convex lens are 10 cm and 20 cm. The refractive index of the glass is 1·5. What is the power of this lens (in units of dioptre) ?

[Q106 · Apr · 2017]

Sign the radii — convex faces are not both positive

For a double convex lens R₁ is positive but R₂ is negative, so 1/R₁ − 1/R₂ becomes 1/R₁ + 1/|R₂| — the terms ADD. Treating both as the same sign halves your answer.

Concept 5 of 6

Lenses in contact — powers add

Intuition

Put two thin lenses against each other and they behave like one lens whose strength is just the sum of the two. Because power adds so cleanly, combination problems are almost always faster in power than in focal length — convert each f to a power, add, then convert back if needed.

Definition

For thin lenses in contact, the powers add:

P=P1+P2+=1f1+1f2+P = P_1 + P_2 + \dots = \dfrac{1}{f_1} + \dfrac{1}{f_2} + \dots
and the combined focal length is f=1/Pf = 1/P.

  • A convex (+) and a concave (−) lens partly cancel; the sign of the net power tells you whether the combination converges (+) or diverges (−).
  • Always work in dioptres (f in metres) — adding focal lengths directly is wrong.

Combination of thin lenses in contact

P=P1+P2,f=1PP = P_1 + P_2, \qquad f = \dfrac{1}{P}
  • P_1, P_2powers of the individual lenses (D)
  • Ppower of the combination (D)
  • ffocal length of the combination (m)

Worked example

A convex lens of power +3 D is placed in contact with a concave lens of power −1 D. Find the focal length of the combination.
  1. Powers add: P=+3+(1)=+2P = +3 + (-1) = +2 D.
  2. Combined focal length f=1/P=1/2=0.5f = 1/P = 1/2 = 0.5 m.
  3. Net power is positive, so the combination is converging.
Answer:+2 D, f = 0.5 m (converging).
Practice this conceptself-check · 4 quick reps

Try it yourself

Two convex lenses of focal lengths 50 cm and 25 cm are placed in contact. Find the net power of the combination.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Two lenses in contact: do focal lengths or powers add?
  2. 2.
    Net power of +2 D and +2 D in contact?
  3. 3.
    Combined focal length when net power is +4 D?
  4. 4.
    Net power of +2.5 D and −2.0 D in contact, and its focal length?

From the bank · past-year question

Example 5Light and OpticsMODERATE
Two convex lenses with power 2 dioptre are kept in contact with each other. The focal length of the combined lens system is

[Q105 · Apr · 2018]

Add powers, not focal lengths

For lenses in contact the POWERS add. Two +2 D lenses give +4 D ⟹ f = 0.25 m, NOT 1 m. Never add focal lengths directly.

Concept 6 of 6

Lens magnification and image formation

Intuition

Magnification for a lens is image distance over object distance — its sign tells you whether the image is upright or inverted, its size tells you how much bigger or smaller. As with mirrors, get the signs from the convention first, then the arithmetic is straightforward.

Definition

Magnification: m=hh=vum = \dfrac{h'}{h} = \dfrac{v}{u} for a lens.

  • m>0m > 0: virtual, erect image. m<0m < 0: real, inverted image.
  • m>1|m| > 1: magnified; m<1|m| < 1: diminished.

Convex-lens image table (object moving in): beyond 2F → real, inverted, diminished (between F and 2F); at 2F → real, inverted, same size; between F and 2F → real, inverted, magnified; at F → at infinity; inside F → virtual, erect, magnified (the magnifying glass).

Lens magnification

m=hh=vum = \dfrac{h'}{h} = \dfrac{v}{u}
  • mmagnification
  • vimage distance
  • uobject distance

Worked example

A convex lens of focal length 15 cm forms an image of an object placed 30 cm away. Find the image distance and magnification.
  1. u=30u = -30 cm, f=+15f = +15 cm.
  2. 1v=1f+1u=115130=230130=130\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{15} - \dfrac{1}{30} = \dfrac{2}{30} - \dfrac{1}{30} = \dfrac{1}{30}, so v=+30v = +30 cm.
  3. m=v/u=30/(30)=1m = v/u = 30/(-30) = -1: real, inverted, same size (object at 2F).
Answer:v=+30v = +30 cm; m=1m = -1 (real, inverted, same size).
Practice this conceptself-check · 4 quick reps

Try it yourself

A concave lens of focal length 10 cm forms an image at a distance of 5 cm from the lens. What is the magnification?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

  1. 1.
    Lens magnification formula?
  2. 2.
    A magnification of −2 for a lens means…
  3. 3.
    Object inside the focus of a convex lens — image is…
  4. 4.
    Object at 2F of a convex lens — image size?

From the bank · past-year question

Example 6Light and OpticsHARD
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?

[Q98 · Apr · 2022]

Lens m = v/u (no minus); mirror m = −v/u

The magnification formula differs by a sign between lens and mirror. For a lens m = v/u; for a mirror m = −v/u. The sign convention then makes both give the right orientation — but use the correct one for the device.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

Watch out for (6)

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 1Light and OpticsMODERATE
Which one among the following figures correctly represents the ray diagram ? (Consider the lens to be thin) (a) Biconvex lens with rays meeting at f on both sides (b) Lens with object at 2f, rays converging past lens (c) Lens with rays from beyond 2f (d) Lens with markers at 2f and f

[Q67 · Sep · 2024]

Example 2Light and OpticsEASY
The focal length of a concave lens is 0.5 m. The power of the lens is

[Q51 · Sep · 2025]

Example 3Light and OpticsEASY
Which one among the following is the correct focal length of a combination of lenses of power 2.5 D and -2.0 D ?

[Q121 · Sep · 2024]

Example 4Light and OpticsHARD
For a human eye, where u is the distance of an object from the eye, f is the focal length of the lens and v is the distance of image from the eye, which is the correct schematic graph ?

[Q128 · Apr · 2024]

Example 5Light and OpticsEASY
If the focal length of a convex lens is 50 cm, which one of the following is its power?

[Q110 · Sep · 2018]

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