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Wize High School Grade 12 Physics Textbook > Geometric Optics

Mirrors Equation

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Mirrors Equation



The distance of the image in relation to the object is described by the mirrors & lenses equation:


 1f=1di+1do \boxed{ \ \dfrac{1}{f}=\dfrac{1}{d_i}+\dfrac{1}{d_o} \ }

  • dod_o is the distance between the object and the mirror
  • did_i is the distance between the image and the mirror
  • ff is the focal length of the mirror

The magnification of the mirror is given by:

 m=−dido=hiho \boxed{ \ m=-\frac{d_i}{d_o}=\frac{h_i}{h_o} \ }
  • mm is the magnification
  • hoh_o is the height of the object
  • hih_i is the height of the image


Note: you might see different notations for the distances ( did_i and dod_o, ii and oo, pp and qq ).



Wize Concept
  • ff is positive for concave mirrors, and negative for convex mirrors
  • dod_o and hoh_o are always positive
  • did_i is positive if the image is in front of the mirror, and negative if behind
  • hih_i is positive if the image is upright, and negative if inverted
  • mm is positive if the image is upright, and negative if inverted
  • real images are inverted, virtual images are upright


Exam Tip
Using these equations we can answer the following questions:
  1. Where is the image?
  2. Is it real or virtual?
  3. Is it upright or inverted?
  4. How big is the image (relative to the object)?

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Example: Convex Mirror


A convex mirror has a focal length of 1010 cm. If the object is placed 2020 cm in front of the mirror, describe the image formed.

Let's use the mirror equation to solve for the image distance did_i:

1f=1di+1do\dfrac{1}{f}=\dfrac{1}{d_i}+\dfrac{1}{d_o}

1di=1f−1do\dfrac{1}{d_i}=\dfrac{1}{f}-\dfrac{1}{d_o}

We have the focal length f=−10f=-10 (negative since the mirror is convex), and the object distance do=20d_o=20. Put these in the equation:

1di=1−10−120=−320\dfrac{1}{d_i}=\dfrac{1}{-10}-\dfrac{1}{20}=-\dfrac{3}{20}

Taking the reciprocal of both sides we get:

di=−203=6.67d_i=-\dfrac{20}{3}=6.67 (cm)

Since the image distance is negative, the image appears to be behind the mirror, and therefore virtual.


Let's find the magnification:

m=−dido=203⋅120=13m=-\dfrac{d_i}{d_o}=\dfrac{20}{3}\cdot\dfrac{1}{20}=\dfrac{1}{3}

Since magnification is also the ratio of the heights, we get:

m=hiho=13   →   hi=13 hom=\dfrac{h_i}{h_o}=\dfrac{1}{3} \ \ \ \to \ \ \ h_i=\dfrac{1}{3} \ h_o

which means that the image is three times smaller than the object.

Also, since hoh_o is always positive, we must have a positive image height hih_i , which means that the image is upright.



Practice: Taller Image


A man standing 2.02.0 m in front of a mirror sees an inverted image of himself 0.480.48 m in front of the mirror.

a) Is the mirror concave or convex?

b) What is the focal length of the mirror?

c) How far from the mirror should he stand if he wants to form an upright image of himself that is 2.52.5 times as high?