Numford
Formula

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

For a mirror, m = h′/h = -v/u. A convex mirror of radius 3.00 m with an object 5.00 m away gives v = +1.15 m and m = +0.23: an erect image about a quarter of the size.

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+1.15 mbehind the mirror
Magnification m+0.23erect, 0.23 times
Image height h′+0.23 mobject +1.00 m
Focal length f+1.50 mR = 2f = +3.00 m
NatureVirtual, erect, diminishedObject between infinity and the pole P
Image positionBetween P and F, behind the mirrorDiminished

Convex mirror of focal length +1.50 m. An object +1.00 m tall at u = -5.00 m: the image is at v = +1.15 m, virtual, erect, diminished.

Unit
m
Show the working
  1. f = R/2 = +3.00/2 = +1.50 m, u = -5.00 m (the object is in front, so u is negative), h = +1.00 m.
  2. 1/v = 1/f - 1/u = 1/(+1.50) - 1/(-5.00) = 6.50/7.50
  3. v = +1.15 m: positive, so the image is behind the mirror and virtual.
  4. m = -v/u = -(+1.15)/(-5.00) = +0.23: positive, so erect, and diminished.
  5. h′ = m × h = +0.23 × +1.00 = +0.23 m
NCERT example

Rear-view mirror, R = 3.00 m, car 5.00 m away

  1. A convex mirror used as a rear-view mirror has a radius of curvature of 3.00 m. A bus is 5.00 m from it. Find the position, nature and size of the image.
  2. f = R/2 = +3.00/2 = +1.50 m, u = -5.00 m (the object is in front, so u is negative), h = +1.00 m.
  3. 1/v = 1/f - 1/u = 1/(+1.50) - 1/(-5.00) = 6.50/7.50
  4. v = +1.15 m: positive, so the image is behind the mirror and virtual.
  5. m = -v/u = -(+1.15)/(-5.00) = +0.23: positive, so erect, and diminished.
  6. h′ = m × h = +0.23 × +1.00 = +0.23 m
Second example

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

  1. An object 2 cm tall is 10 cm in front of a concave mirror of focal length 15 cm.
  2. f = R/2 = -30/2 = -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) = 5/150 = 1/30
  4. v = +30 cm: positive, so the image is behind the mirror and virtual.
  5. m = -v/u = -(+30)/(-10) = +3: positive, so erect, and enlarged.
  6. h′ = m × h = +3 × +2 = +6 cm
Reading m

What the sign and size tell you

  • m is positive for an erect image and negative for an inverted one.
  • For a mirror, a negative m always goes with a real image and a positive m with a virtual image.
  • |m| more than 1 means the image is enlarged, less than 1 diminished, and exactly 1 the same size.
  • A convex mirror always gives m between 0 and +1.
Questions

Quick answers

What is the magnification formula for a mirror?

m = h′/h = -v/u, where h′ is the image height, h the object height, and u and v the object and image distances with their signs.

Why is there a minus sign in m = -v/u for a mirror?

For a real image, u and v are both negative, so v/u is positive, yet the image is inverted. The minus sign makes m negative for a real, inverted image.

What does m = +0.23 mean?

The image is erect (positive) and 0.23 times the size of the object, as for the rear-view mirror in the NCERT example.

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