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Magnification of Lens Formula

Waves, sound and optics formulasNo unitClass 10Updated 1 Nov 2026

m=h′h=vum = \frac{h'}{h} = \frac{v}{u}
SI unit
no unit
Dimensions
[M⁰ L⁰ T⁰]
where
  • m magnification
  • h′ image height m
  • h object height m
  • v image distance m
  • u object distance m

The magnification of lens formula is m=h′h=vum = \frac{h'}{h} = \frac{v}{u}, image height over object height, which equals image distance over object distance. Unlike the mirror version, there is no minus sign. A result of m = −3 tells you the image is three times taller than the object and inverted, while m = +0.25 means an upright image a quarter of the size.

Lenses form real images on the far side from the object, so v and u then have opposite signs in the New Cartesian convention and v/u is already negative. That is why the lens formula for magnification needs no extra sign. Magnification carries no unit at all. It is the second step in nearly every lens numerical in Class 10 and Class 12: find v with the lens formula, then size the image with m.

Quick facts

  • Formula

    m=h′h=vum = \frac{h'}{h} = \frac{v}{u}

  • Using focal length

    m=ff+u=f−vfm = \frac{f}{f + u} = \frac{f - v}{f}

  • Unit

    None

  • Dimensional formula

    [M⁰ L⁰ T⁰]

  • Negative m

    Real, inverted image on the other side of the lens

  • Positive m

    Virtual, upright image on the same side as the object

  • Concave lens

    m is always between 0 and +1

Magnification formula of lens, term by term

Signs used for a thin lens
SymbolStands forSign in New Cartesian convention
mmLinear magnificationNegative when the image is upside down
h′h'Image heightPositive if the image points up
hhObject heightTaken as positive
vvDistance of the image from the optical centrePositive on the outgoing side of the lens
uuDistance of the object from the optical centreNegative, light arrives from that side

Distances are measured from the optical centre O. Light is drawn travelling left to right, so anything to the right of the lens has a positive distance and anything to the left a negative one.

Deriving m = v/u for a lens

A ray leaving the tip of the object and heading through the optical centre passes straight on without bending. It forms two similar triangles: one with the object height h and the distance |u|, the other with the image height h′ and the distance |v|. So in size ∣h′∣h=∣v∣∣u∣\frac{|h'|}{h} = \frac{|v|}{|u|}.

For a real image from a convex lens, h′ is negative, v is positive and u is negative. The ratio v/u is negative, exactly matching the sign of h′/h. No correction is needed, so m=vum = \frac{v}{u} holds as it stands.

Eliminating v with the thin lens relation 1v−1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f} gives v=ufu+fv = \frac{uf}{u + f}. Divide by u:

m=vu=ff+um = \frac{v}{u} = \frac{f}{f + u}

Solved questions on the magnification of lens formula

Example 1: a convex lens projecting a real image

A candle is 18 cm from a convex lens of focal length 12 cm. Find where the image forms and how magnified it is.

u = −18 cm, f = +12 cm.

1v=112−118=3−236=136⇒v=+36 cm\frac{1}{v} = \frac{1}{12} - \frac{1}{18} = \frac{3 - 2}{36} = \frac{1}{36} \Rightarrow v = +36\ \text{cm}

m=vu=36−18=−2m = \frac{v}{u} = \frac{36}{-18} = -2

Real, inverted and twice the candle's height, on the far side of the lens. Check: f/(f + u) = 12/(−6) = −2.

Example 2: a diverging lens

A concave lens with f = −20 cm views a toy 30 cm away. Find m.

1v=−130−120=−2+360⇒v=−12 cm\frac{1}{v} = -\frac{1}{30} - \frac{1}{20} = -\frac{2 + 3}{60} \Rightarrow v = -12\ \text{cm}

m=−12−30=+0.40m = \frac{-12}{-30} = +0.40

The image is virtual, upright and smaller, on the same side as the toy.

Example 3: a projector, mixing cm and m

A slide 2.4 cm tall sits 10 cm from a projector lens, and a sharp image appears on a screen 5.0 m away. How tall is the picture on the screen?

Use the same unit for both distances: u = −0.10 m, v = +5.0 m.

m=5.0−0.10=−50m = \frac{5.0}{-0.10} = -50

h′=mh=−50×2.4 cm=−120 cm=−1.2 mh' = m h = -50 \times 2.4\ \text{cm} = -120\ \text{cm} = -1.2\ \text{m}

The picture is 1.2 m tall and upside down, which is why slides are loaded inverted.

Example 4: a magnifying glass

A reading lens has f = +5.0 cm and is held 4.0 cm above a printed letter. Find the magnification.

m=ff+u=5.05.0−4.0=+5.0m = \frac{f}{f + u} = \frac{5.0}{5.0 - 4.0} = +5.0

Upright and five times bigger. The object sits inside the focal length, so the image is virtual.

Sign of m for different object positions

Convex lens images and their magnification
Object positionImagem
Beyond 2FReal, inverted, smaller, between F and 2FBetween −1 and 0
At 2FReal, inverted, same size, at 2F−1
Between F and 2FReal, inverted, larger, beyond 2FLess than −1
Inside FVirtual, upright, larger, same sideGreater than +1

A concave lens never forms anything but a virtual, upright, reduced image, so its magnification always falls between 0 and +1. If a concave lens problem gives you m = −2, a sign has slipped somewhere.

Mistakes and linked formulas

  • Borrowing the mirror minus sign. The magnification formula of mirror has m = −v/u; the lens version does not.
  • Mixing units for u and v. They must be in the same unit, though the unit itself cancels.
  • Confusing linear magnification with power. A lens of high power does not automatically give a large m; that depends on where the object is.
  • Skipping the lens formula. You usually need v first from the formula of lens.

Two lenses in a row multiply their magnifications: m=m1×m2m = m_1 \times m_2. That is how compound microscopes reach large values. The focal length itself comes from the lens maker formula, and lens power is just 1/f. See the waves and optics hub for the full chapter.

Frequently asked questions

What is the formula for magnification of lens?

Divide v by u, or the image height by the object height: m = v/u = h′/h. Written with the focal length, m = f/(f + u).

Why is there no minus sign in the lens magnification formula?

Light crosses a lens, so a real image lands on the far side from the object. The two distances then carry opposite signs and v/u turns negative without help, which already marks the image as inverted. A mirror sends light back, so its real images share the object's side and need the extra minus.

Is the magnification of a concave lens always positive?

Yes. A concave lens on its own always forms a virtual, upright, diminished image, so m lies between 0 and +1.

What does m = −1 mean for a convex lens?

The image is real, inverted and the same size as the object. This happens when the object is at 2F, a distance of twice the focal length.

Does lens magnification have a unit?

No. Both quantities in the ratio are lengths, so the metres cancel and m is just a number.