Numford
Formula

Magnification formula for a lens: m = h′/h = v/u

For a lens, m = h′/h = v/u. A convex lens with f = 10 cm and an object at 15 cm gives m = -2, a real image twice the size and upside down.

The ray diagram

Drag the object or type values. Signs follow the New Cartesian convention: distances are measured from the pole or optical centre, along the incident light is positive, against it negative, and heights above the axis are positive. So u is negative for a real object.

Image distance v+30 cmbehind the lens
Magnification m-2inverted, 2 times
Image height h′-4 cmobject +2 cm
Power P+10 Dconverging
NatureReal, inverted, enlargedObject between F₁ and 2F₁
Image positionBeyond 2F₂Enlarged

Convex lens of focal length +10 cm. An object +2 cm tall at u = -15 cm: the image is at v = +30 cm, real, inverted, enlarged.

Unit
cm
Show the working
  1. f = +10 cm, u = -15 cm (the object is in front, so u is negative), h = +2 cm.
  2. 1/v = 1/f + 1/u = 1/(+10) + 1/(-15) = 5/150 = 1/30
  3. v = +30 cm: positive, so the image is behind the lens and real.
  4. m = v/u = (+30)/(-15) = -2: negative, so inverted, and enlarged.
  5. h′ = m × h = -2 × +2 = -4 cm
  6. Power P = 1/f = 1/(+0.1 m) = +10 D
Solved example 1

Convex lens, f = 10 cm, object at 15 cm

  1. An object 2 cm tall is 15 cm from a convex lens of focal length 10 cm.
  2. f = +10 cm, u = -15 cm (the object is in front, so u is negative), h = +2 cm.
  3. 1/v = 1/f + 1/u = 1/(+10) + 1/(-15) = 5/150 = 1/30
  4. v = +30 cm: positive, so the image is behind the lens and real.
  5. m = v/u = (+30)/(-15) = -2: negative, so inverted, and enlarged.
  6. h′ = m × h = -2 × +2 = -4 cm
  7. Power P = 1/f = 1/(+0.1 m) = +10 D
Solved example 2

Concave lens, f = 15 cm, object at 10 cm

  1. An object 2 cm tall is 10 cm from a concave lens of focal length 15 cm.
  2. f = -15 cm, u = -10 cm (the object is in front, so u is negative), h = +2 cm.
  3. 1/v = 1/f + 1/u = 1/(-15) + 1/(-10) = -25/150 = -1/6
  4. v = -6 cm: negative, so the image is on the object side and virtual.
  5. m = v/u = (-6)/(-10) = +0.6: positive, so erect, and diminished.
  6. h′ = m × h = +0.6 × +2 = +1.2 cm
  7. Power P = 1/f = 1/(-0.15 m) = -6.67 D
Reading m

What the sign and size tell you

  • m is positive for an erect, virtual image and negative for an inverted, real image.
  • |m| more than 1 means enlarged, less than 1 diminished, and exactly 1 the same size.
  • A concave lens always gives m between 0 and +1.
Questions

Quick answers

What is the magnification formula for a lens?

m = h′/h = v/u, where h′ and h are the image and object heights and u and v the object and image distances with their signs.

Why is there no minus sign for a lens?

For a real image u is negative and v positive, so v/u is already negative, which matches an inverted image.

What is the magnification for an object 15 cm from a convex lens of focal length 10 cm?

v = +30 cm, so m = v/u = +30/(-15) = -2. The image is real, inverted and twice as large.

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