Lesson 4 of 10Light & Geometric Optics

Unit 4 · Lesson 4 of 10

The Mirror and Magnification Equations

Turning ray diagrams into numbers: the mirror equation, the magnification equation, and the real-is-positive sign convention that tells you whether an image is real or virtual, upright or inverted.

3 min read11 practice questionsInteractiveAsk AI about this →

Key things to remember

1Mirror equation: $\frac{1}{f} = \frac{1}{d _{o}} + \frac{1}{d _{i}}$
2Magnification: $M = \frac{h _{i}}{h _{o}} = - \frac{d _{i}}{d _{o}}$
3 $f$ is positive for concave mirrors and negative for convex mirrors.
4 $d_{i}$ positive means a real image in front of the mirror; $d_{i}$ negative means a virtual image behind it.
5 $M$ negative means inverted; $M$ positive means upright. Its size beyond or below 1 tells larger or smaller.

Ray diagrams show you roughly where an image is and what it looks like. But an engineer designing a car headlight or a telescope needs exact distances, and a careful ray diagram takes time. Two short equations do the same job in a few lines of arithmetic. The only new skill is reading the signs: a minus sign in the answer is not an error, it is information about the image.

\frac{1}{f} = \frac{1}{d _{o}} + \frac{1}{d _{i}}

The mirror equation: f is the focal length, d_o is the object distance and d_i is the image distance, all measured from the vertex.

M = \frac{h _{i}}{h _{o}} = - \frac{d _{i}}{d _{o}}

The magnification equation: image height over object height, which also equals minus image distance over object distance.

The sign convention (real is positive)

Quantity	Positive (+) when	Negative (−) when
Focal length $f$	Concave (converging) mirror	Convex (diverging) mirror
Object distance $d_{o}$	Real object in front of the mirror (every case in this course)	Not used in this course
Image distance $d_{i}$	Real image, in front of the mirror	Virtual image, behind the mirror
Image height $h_{i}$	Upright image	Inverted image
Magnification $M$	Upright image	Inverted image

Object heights are taken as positive, so the sign of the image height matches the sign of M.

Reading the magnification

The size of M (ignoring its sign) tells you the size: greater than 1 means larger, exactly 1 means the same size, less than 1 means smaller. The sign of M tells you the attitude: negative means inverted, positive means upright. So

M = - 0.50

is an inverted image half the size of the object.

Fig 1The equations reproduce the ray diagram: with the object beyond C, the mirror equation gives a positive image distance between F and C and a negative magnification whose size is less than 1, a real, inverted, smaller image.

A method that works every time

List the givens with signs: is the mirror concave ( $f$ positive) or convex ( $f$ negative)?
Rearrange the mirror equation for the unknown before substituting, for example $\frac{1}{d _{i}} = \frac{1}{f} - \frac{1}{d _{o}}$ .
Calculate the reciprocal, then flip it: the last step is always 'one over' your result.
Interpret the sign of $d_{i}$ : positive is real (in front), negative is virtual (behind).
Find $M = - \frac{d _{i}}{d _{o}}$ , then $h_{i} = M h_{o}$ , and state SALT.

Do not forget to flip

Solving gives

\frac{1}{d _{i}}

, not

d_{i}

. If your calculation shows

\frac{1}{d _{i}} = 0.0667

, the image distance is

\frac{1}{0.0667} = 15 cm

, not

0.0667 cm

. A quick check: image distances should be similar in size to the other distances in the problem.

The equations agree with every row of the concave-mirror table from the last lesson. Put the object at C ( $d_{o} = 2 f$ ) and the equation gives $d_{i} = 2 f$ and $M = - 1$ : same size, inverted, at C. Put the object inside F ( $d_{o} < f$ ) and $\frac{1}{f} - \frac{1}{d _{o}}$ becomes negative, so $d_{i}$ is negative: a virtual image. Put it exactly at F and $\frac{1}{d _{i}} = 0$ , which has no solution: no image forms. For a convex mirror, $f$ is negative, so $d_{i}$ always comes out negative and $M$ positive and less than 1, exactly as the ray diagrams showed.

Loading interactive…

Drag the object in front of a concave or convex mirror and follow the mirror equation solved step by step, with the signs of the image distance and magnification explained, while the rays show where the image forms.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

4.0 cm

tall candle stands

30.0 cm

from a concave mirror with focal length

10.0 cm

. Find the image distance, the magnification and the image height, and describe the image.

Example 2

A face is

8.0 cm

from a concave makeup mirror with

f = 12.0 cm

. Find the image distance and magnification. If a freckle is

2.0 mm

wide, how wide does its image appear?

Example 3

A security mirror is convex with a focal length of magnitude

15.0 cm

. A shopper's face is

30.0 cm

away. Find the image distance and magnification.

Where marks get lost

Forgetting to take the reciprocal at the end, and reporting $\frac{1}{d _{i}}$ as the image distance.
Using a positive focal length for a convex mirror. Convex mirrors and diverging lenses have negative $f$ .
Adding the reciprocals instead of subtracting: $\frac{1}{d _{i}} = \frac{1}{f} - \frac{1}{d _{o}}$ , not $\frac{1}{f} + \frac{1}{d _{o}}$ .
Dropping the minus sign in $M = - \frac{d _{i}}{d _{o}}$ , which flips every upright/inverted conclusion.
Calling $1/ (a + b)$ the same as $1/ a + 1/ b$ . You cannot flip each term separately; combine the fractions first, then flip.

Flashcards 7

Mirror equation

Magnification equation

Sign of f for mirrors

Negative image distance

+ 3 more

Practice

An object is

60.0 cm

from a concave mirror with a focal length of

20.0 cm

. Find the image distance, in cm, to 1 decimal place.

For the mirror in the previous question (

d_{o} = 60.0 cm

d_{i} = 30.0 cm

), what is the magnification? Include the sign, to 2 decimal places.

An object is

10.0 cm

from a concave mirror with

f = 15.0 cm

. Find the image distance in cm, including its sign, to 1 decimal place.

A convex mirror has

f = - 20.0 cm

. An object is

20.0 cm

in front of it. Find the magnification, to 2 decimal places.

4.0 cm

tall object is

25.0 cm

from a concave mirror with

f = 10.0 cm

. Find the image height in cm, including its sign, to 1 decimal place.

A mirror forms a real image

24.0 cm

from it when the object is

40.0 cm

away. Find the focal length in cm, to 1 decimal place.

A concave mirror with

f = 12.0 cm

forms a real image

36.0 cm

in front of it. How far is the object from the mirror, in cm, to 1 decimal place?

A calculation gives

d_{i} = - 6.0 cm

and

M = + 0.40

. Which description is correct?

Using the mirror equation, what happens when an object is placed exactly at the focal point of a concave mirror?

Match each mirror result to what it tells you.

d_{i} = + 15 cm

d_{i} = - 24 cm

M = - 0.50

M = + 3.0

f = - 15 cm

A calculation gives a magnification of

M = - 2.0

. Is the image upright or inverted? (One word.)

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