Starting from the formula for refraction at a single spherical surface applied in turn to the two surfaces of a thin lens, derive the thin lens formula $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$.

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Concept understanding — Thin Lens Formula

The Thin Lens Formula: Why a Lens Behaves the Way It Does

When you hold a magnifying glass and move it toward a piece of paper, something dramatic happens. At first the image is blurry, then suddenly a sharp, bright spot appears — and if you hold it still, the paper can catch fire. That spot is the image of the sun, and the distance from the lens to the paper is the focal length. The thin lens formula is the mathematical rule that governs exactly where that image will form for any object you place in front of the lens.

The Intuition: Bending Light Systematically

A thin lens works by bending every ray of light that passes through it. The key idea is that the bending is predictable — it depends only on where the ray hits the lens and the lens's own power (its focal length). For a thin lens, we assume the lens is so thin that we can ignore its thickness and treat all bending as happening at a single plane through its centre.

Imagine an object placed at some distance $u$ from the lens. Light from each point on the object spreads out in all directions. The lens intercepts this light and redirects it so that all rays from a single point on the object meet again at a single point on the other side — that meeting point is the image. The distance from the lens to that image is $v$ .

The relationship between $u$ , $v$ , and the focal length $f$ is not arbitrary. It comes from geometry: similar triangles formed by the rays and the lens surface give a clean, simple equation.

The Precise Statement

For a thin lens, the object distance $u$ , image distance $v$ , and focal length $f$ are related by:

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

This is the thin lens formula in its Cartesian sign convention form.

Watch out

The sign convention is everything

In the Cartesian convention (used in most Indian board exams):

Distances measured against the direction of incident light are negative.
Distances measured along the direction of incident light are positive.
For a convex lens, $f$ is positive; for a concave lens, $f$ is negative.
$u$ is always negative (object is on the incident side).
$v$ is positive for a real image (on the opposite side) and negative for a virtual image (on the same side as the object).

If you use the older "real is positive" convention, the formula looks like $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ . The physics is identical — only the signs change. Stick to one convention and use it consistently.

What the Formula Tells You

The formula says that the curvature of the wavefront (the reciprocal of distance) changes by a fixed amount as light passes through the lens. That fixed amount is $1/f$ , the lens's power measured in dioptres.

If the object is very far away ( $u \to -\infty$ ), then $1/u \to 0$ , so $1/v = 1/f$ , meaning $v = f$ . The image forms at the focal point — this is why you can burn paper with sunlight.
If the object is at twice the focal length ( $u = -2f$ ), then $1/v = 1/f + 1/(-2f) = 1/(2f)$ , so $v = 2f$ . The image is the same size as the object and inverted. …

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For a thin lens, light first refracts at surface 1 (radius $R_1$ ), forming an intermediate (possibly virtual) image at some distance $v_1$ from the (common, since the lens is thin) optical centre; applying the single-surface formula to this first refraction (with $n_1=1$ for the surrounding air and $n_2=n$ for the lens material) gives $\dfrac{n}{v_1}-\dfrac{1}{u}=\dfrac{n-1}{R_1}$ . This intermediate image then acts as the object for the second surface (radius $R_2$ ), where light refracts back out of the lens into air; applying the single-surface formula again for this second refraction (now $n_1=n$ , $n_2=1$ , and the object distance for this surface is $v_1$ , the first surface's image distance) gives $\dfrac{1}{v}-\dfrac{n}{v_1}=\dfrac{1-n}{R_2}$ . Adding these two equations, the $n/v_1$ term appears with opposite signs in the two equations and cancels exactly, leaving $\dfrac1v-\dfrac1u=(n-1)\left(\dfrac1{R_1}-\dfrac1{R_2}\right)$ . Since the right-hand side, for a fixed lens, is a fixed constant depending only on the lens's material and s …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 18 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. The diagram depicts a$
(a) concave lens
(b) convex lens
(c) plane sheet of glass since it is of equal thickness
(d) convexo-concave lens
›Reveal solutionSolution
The sketched cross-section — thin at the centre, thicker towards the edges — is the standard simplified symbol for a diverging (concave) lens, as opposed to a convex lens (thick at the centre) or a plane sheet of glass (uniform thickness with flat, parallel faces).

Reasoning from the shape of the diagram

Lenses are conventionally distinguished in simplified textbook cross-section diagrams by how their thickness varies from the optical centre to the edge:

A convex (converging) lens bulges outward: it is thickest at the centre and thinner at the edges.

A concave (diverging) lens curves inward at both faces: it is thinnest at the centre and thicker at the edges (a crescent-like shape when only the outline is sketched).

A plane sheet of glass has two flat, parallel faces and therefore uniform thickness everywhere. …
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CBSE 2024Set IMPROVEMENT1 markMCQ
$f$
(a) $F<f$
(b) $F>f$
(c) $F=\infty$
(d) $F=f$
›Reveal solutionSolution

Two lenses of focal length $f$ each, in contact, combine to give $F=f/2$ , which is less than $f$ .

…

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CBSE 2024Set ANNUAL1 mark
$Q. For a convex lens, in which position an object should be placed so as to have real and magnified image?$
›Reveal solutionSolution
A convex lens forms a real, magnified image only when the object lies between F and 2F.

For a thin convex (converging) lens, the nature, size and position of the image depend on where the object is placed relative to the focal point F and the point 2F:

Object beyond 2F → real, inverted, DIMINISHED image between F and 2F on the other side.

Object exactly at 2F → real, inverted image of the SAME size, also at 2F.

Object between F and 2F → real, inverted, MAGNIFIED image formed beyond 2F on the other side.

Object at F → image at infinity (highly magnified, not usable as a real finite image).

Object between the lens and F → virtual, erect, magnified image on the same side (used as a simple magnifier). …
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CBSE 2023Set ANNUAL1 markMCQ
$Q. The focal length of concave lens is 20 cm. Its power is :$
(a) - 0.2 D
(b) + 5 D
(c) - 5 D
(d) infinity
›Reveal solutionSolution

A concave lens of focal length 20 cm has power -5 D.

Power P = 1/f, with f in metres. For a concave (diverging) lens the focal length is negative: f = -20 cm = -0.20 m.

P = 1/(-0.20) = -5 D. …

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CBSE 2022Set ANNUAL1 mark
$Q. Two thin lenses of power +5 Dioptre and -3 Dioptre are placed in contact. Find the power of this combination.$
›Reveal solutionSolution

For thin lenses in contact the powers add: $P = 5 + (-3) = +2$ D.

When two thin lenses are placed in contact, the equivalent power is the algebraic sum of the individual powers:

$P = P_1 + P_2.$

Here $P_1 = +5$ D (convex) and $P_2 = -3$ D (concave):

$P = (+5) + (-3) = +2 \text{ D}.$

…

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CBSE 2020Set XS1 mark
$Q. Define second focus of a concave lens.$
›Reveal solutionSolution

It is the virtual point from which parallel incident rays seem to diverge after passing through the diverging lens.

Definition. The second principal focus $F_2$ of a lens is the point associated with rays that strike the lens parallel to the principal axis.

For a concave (diverging) lens, such parallel rays diverge after refraction. When produced backwards they appear to meet at a point on the principal axis on the same side as the incident light. That point is the second focus $F_2$ ; it is a virtual focus, and the focal length is take …

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CBSE 2019Set ANNUAL1 mark
$Q. For which position of the object magnification of convex lens is -1. (minus one)?$
›Reveal solutionSolution

$m = -1$ for a convex lens occurs when the object is at a distance $2f$ from the lens; the image is then real, inverted and equal in size.

Concept. Linear magnification of a lens is $m = \dfrac{v}{u}$ . A value $m = -1$ means the image is the same size as the object but inverted (real image).

…

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CBSE 2019Set ANNUAL1 mark
$Q. Write the SI unit for power of a lens.$
›Reveal solutionSolution

SI unit of lens power = dioptre (D) = $\text{m}^{-1}$ .

Concept. The power of a lens measures its ability to converge or diverge light and is defined as the reciprocal of its focal length expressed in metres:

$P = \frac{1}{f\,(\text{in metres})}$

…

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CBSE 2019Set ANNUAL1 markMCQ
$Q. Cylindrical lenses are used to correct the eye defect called -$
(a) myopia
(b) hypermetropia
(c) astigmatism
(d) presbyopia
›Reveal solutionSolution

Cylindrical lenses correct astigmatism.

In astigmatism the cornea (or lens) is not equally curved in all planes, so it focuses horizontal and vertical lines differently and the eye cannot see both sharply at once. A cylindrical lens has power in only one plane, so it can compensate for the unequal curvature. …

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CBSE 2019Set ANNUAL1 markMCQ
$Q. Equivalent focal length of two lenses in contact having power - 15 D and + 5 D will be -$
(a) - 20 cm
(b) - 10 cm
(c) + 10 cm
(d) + 20 cm
›Reveal solutionSolution

P = P₁ + P₂ = −15 + 5 = −10 D ⇒ f = 1/P = −10 cm.

For thin lenses in contact the net power is the algebraic sum of individual powers:

$P = P_1 + P_2 = (-15) + (+5) = -10\ \text{D}.$

The equivalent focal length is

…

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CBSE 2019Set ANNUAL1 mark
$Q. Write relation between power of lens and it's focal length.$
›Reveal solutionSolution
Power of a lens P = 1/f, with f in metres and P in dioptres.

The power of a lens measures its ability to converge or diverge light. It is defined as the reciprocal of the focal length:

P = 1/f

where f is in metres and P is in dioptre (D).

A converging (convex) lens has positive f, so positive power. …
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CBSE 2019Set ANNUAL1 mark
$Q. Define first focus of a concave lens.$
›Reveal solutionSolution

First focus of a concave lens = the point toward which incident rays must head so that refracted rays become parallel to the axis (a virtual point).

The first (or object-side) focal point of a lens is defined by the behaviour of rays that emerge parallel to the principal axis.

For a converging lens it is the point on the axis from which rays diverge to become parallel after refraction. For a concave (diverging) lens, incident rays cannot actually pass through a real focus; instead, the first focus is the point on the principal axis toward which the incident rays must be aimed (converging toward it) so that after refraction they travel parallel to the principal axis.

…

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Q.Starting from the formula for refraction at a single spherical surface applied in turn to the two surfaces of a thin lens, derive the thin lens formula 1v−1u=1f\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}v1​−u1​=f1​.

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Q.Starting from the formula for refraction at a single spherical surface applied in turn to the two surfaces of a thin lens, derive the thin lens formula $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$ .

Unlock everything free for 14 days