A convex lens of focal length $10\ ext{cm}$ forms a real image twice the size of the object. Find the object distance and the image distance.

Question

A convex lens of focal length $10\ 	ext{cm}$ forms a real image twice the size of the object. Find the object distance and the image distance.

Accepted Answer

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.
If the object is between the lens and the focal point ( $|u| < f$ ), then $1/v$ becomes negative, meaning $v$ is negative — the image is virtual, upright, and on the same side as the object. This is how a magnifying glass works.

Tip

A quick check for convex lenses

For a convex lens, if the object is beyond the focal point, the image is real and inverted. If the object is inside the focal point, the image is virtual and upright. The formula automatically gives you the correct sign for $v$ — let the algebra do the work.

Why "Thin"?

The formula assumes the lens is thin enough that we don't need to track where inside the lens the bending happens. For thick lenses or compound lens systems, the formula becomes more complicated (using principal planes). But for a single lens you hold in your hand, the thin lens formula is accurate to within a few percent — and it's all you need for the board exam.

The Core Idea in One Sentence

The thin lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ is a geometric consequence of the fact that a lens bends light by an amount proportional to the distance from its centre, and it tells you exactly where an image will form for any given object distance and focal length.

The thin lens formula is a cornerstone of the NCERT Class 12 Physics Ray Optics and Optical Instruments chapter, and 'thin lens formula 1/v - 1/u = 1/f' or 'lens formula important questions class 12 physics' are among the most searched terms for board revision. This sign-convention-based approach is also essential for nearly every JEE Main, NEET and state CET optics numerical.

For a real, inverted image formed by a convex lens, the magnification $m=v/u$ is negative, so here $m=-2\Rightarrow v=-2u$ . Substituting into the thin lens formula $\dfrac1v-\dfrac1u=\dfrac1f$ : $\dfrac1{-2u}-\dfrac1u=\dfrac1{10}\Rightarrow-\dfrac1{2u}-\dfrac1u=\dfrac1{10}\Rightarrow-\dfrac{1+2}{2u}=\dfrac1{10}\Rightarrow-\dfrac3{2u}=\dfrac1{10}\Rightarrow u=-15\ \text{cm}$ . Then $v=-2u=-2\times(-15)=+30\ \text{cm}$ . Checking: $\dfrac1{30}-\dfrac1{-15}=\dfrac1{30}+\dfrac2{30}=\dfrac3{30}=\dfrac1{10}=\dfrac1f$ ✓.

✓Final answer

u=-15\ \text{cm}

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.

A convexo-concave (meniscus) lens has one face curving outward and the other curving inward, giving a shape that is thicker on one side than uniformly across, but is neither symmetric like a convex nor a concave lens, and is not what a simple two-line crescent sketch depicts.

The figure described is exactly the standard simplified biconcave lens outline (a thin crescent, curving inward on both faces), so it depicts a concave (diverging) lens, not a convex lens, plane sheet, or meniscus lens.

✓Final answer
$(a) concave lens$
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 $f$ .

For two thin lenses of focal lengths $f_1$ and $f_2$ placed in contact, the combined focal length $F$ is given by $\dfrac{1}{F}=\dfrac{1}{f_1}+\dfrac{1}{f_2}$ . Here $f_1=f_2=f$ , so $\dfrac{1}{F}=\dfrac{1}{f}+\dfrac{1}{f}=\dfrac{2}{f}$ , giving $F=\dfrac{f}{2}$ . Since $F=f/2$ is smaller in magnitude than $f$ , we have $F<f$ .

✓Final answer
$F<f$
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).

So for a REAL and MAGNIFIED image, the object must be placed between the focus and 2F, i.e. its distance u from the lens satisfies f < u < 2f. This is exactly the working range used in a slide/film projector, where the object (film) is kept just beyond F to throw a large real image on a distant screen.

✓Final answer
$Between F and 2F (f < u < 2f) — the object distance must lie strictly between one and two focal lengths from the lens.$
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.

The negative sign correctly marks it as a diverging lens — a point stressed in NCERT/CBSE Class 12 optics.

✓Final answer
$(c) -5 D.$
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):

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

The positive value means the combination behaves as a converging lens of focal length $f = 1/P = +0.5$ m $= 50$ cm.

✓Final answer
$+2$
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 taken as negative.

✓Final answer
$The second focus of a concave lens is the point on the principal axis from which a beam of light incident parallel to the axis appears to diverge after refraction — a virtual focus lying on the side of incidence.$
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).

Analysis. Using the lens formula $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ with $m = v/u = -1$ gives $v = -u$ . Substituting: $\dfrac{1}{-u} - \dfrac{1}{u} = \dfrac{1}{f} \Rightarrow -\dfrac{2}{u} = \dfrac{1}{f} \Rightarrow u = -2f$ . So the object lies at $2f$ and the image forms at $2f$ on the other side.

✓Final answer
$2f$
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:

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

The SI unit is therefore $\text{metre}^{-1}$ , given the special name dioptre (D). A lens of $1\,\text{D}$ has focal length $1\,\text{m}$ ; convex lenses have positive power, concave lenses negative.

✓Final answer
$1\,\text{D} = 1\,\text{m}^{-1}$
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. Myopia/hypermetropia are corrected by concave/convex spherical lenses, and presbyopia by bifocal lenses.

✓Final answer
$(c) astigmatism.$
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

$f = \frac{1}{P} = \frac{1}{-10}\ \text{m} = -0.10\ \text{m} = -10\ \text{cm}.$

The negative sign means the combination behaves as a diverging (concave) lens.

✓Final answer
$(b) - 10 cm.$
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.

A diverging (concave) lens has negative f, so negative power.

Example: a convex lens of f = 0.5 m has P = 1/0.5 = +2 D.

✓Final answer
$P = 1/f (f measured in metres), measured in dioptre.$
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.

Because the rays only appear to be heading to this point (they are refracted before reaching it), the first focus of a concave lens is a virtual point, lying on the same side as the incoming light.

✓Final answer
$It is the point on the axis toward which incident rays must be directed so that the refracted rays become parallel to the axis — a virtual point for a concave lens.$

> [!ANSWER]
> $u=-15\ 	ext{cm}$ (object $15\ 	ext{cm}$ in front of the lens), $v=+30\ 	ext{cm}$ (real image $30\ 	ext{cm}$ behind the lens).

Q.A convex lens of focal length 10 cm10\ \text{cm}10 cm forms a real image twice the size of the object. Find the object distance and the image distance.

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Q.A convex lens of focal length $10\ \text{cm}$ forms a real image twice the size of the object. Find the object distance and the image distance.

Unlock everything free for 14 days