Mirror Formula Derivation for a Convex Mirror | Class 10

Quick answer: For a convex mirror, the triangles formed by the object, the image and the centre of curvature are similar, and so are the triangles formed with the pole. Equating the two ratios and using the sign convention ($u$ negative, $v$, $f$ and $R = 2f$ positive) gives $uf + vf = uv$, which simplifies to the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.

Syllabus note: The rationalised NCERT Class 10 book states the mirror formula and asks you to use it in numericals, but it does not ask you to derive it. The derivation below is enrichment, useful for understanding and for Class 12 (Ray Optics). Do not expect it as a Class 10 board question.

The setup

Take a convex mirror of small aperture with pole $P$, principal focus $F$ and centre of curvature $C$. For a convex mirror, $F$ and $C$ lie behind the mirror.

An object $AB$ of height $h$ stands on the principal axis in front of the mirror. A convex mirror always forms a virtual, erect and diminished image $A'B'$, behind the mirror, between $P$ and $F$.

  A                     
  |       mirror        
  |         )  A'       
  |         )  |        
--B---------P--B'--F--C-
  front     )   behind  

The image is found with two rays from $A$:

• Ray 1: A ray towards $C$ hits the mirror along the normal and is reflected back on itself. Extended behind the mirror, it passes through $C$.
• Ray 2: A ray striking the pole $P$ is reflected so that the angle of reflection equals the angle of incidence.

Extended behind the mirror, the reflected rays meet at $A'$, the top of the virtual image.

Derivation of the mirror formula

Step 1: First pair of similar triangles.

In triangles $ABC$ and $A'B'C$:

• $\angle ABC = \angle A'B'C = 90^\circ$
• $\angle ACB = \angle A'CB'$ (common angle at $C$)

So, by AA similarity, triangle $ABC \sim$ triangle $A'B'C$, and:

$\frac{AB}{A'B'} = \frac{BC}{B'C} \qquad \text{(1)}$

Step 2: Second pair of similar triangles.

By the law of reflection, the ray from $A$ to $P$ and the reflected ray (which appears to come from $A'$) make equal angles with the principal axis. In triangles $ABP$ and $A'B'P$:

• $\angle ABP = \angle A'B'P = 90^\circ$
• $\angle APB = \angle A'PB'$

So triangle $ABP \sim$ triangle $A'B'P$, and:

$\frac{AB}{A'B'} = \frac{PB}{PB'} \qquad \text{(2)}$

Step 3: Equate (1) and (2).

$\frac{BC}{B'C} = \frac{PB}{PB'} \qquad \text{(3)}$

Step 4: Write the lengths along the axis from the pole.

From the figure, $BC = PB + PC$ and $B'C = PC - PB'$. Put these in (3):

$\frac{PB + PC}{PC - PB'} = \frac{PB}{PB'} \qquad \text{(4)}$

Step 5: Apply the New Cartesian sign convention.

Distances are measured from $P$; the incident light travels from left to right.

• $PB = -u$ (object is in front, so $u$ is negative)
• $PB' = v$ (image is behind, so $v$ is positive)
• $PC = R = 2f$ (centre of curvature is behind, so $R$ and $f$ are positive)

Equation (4) becomes:

$\frac{-u + 2f}{2f - v} = \frac{-u}{v}$

Step 6: Cross-multiply and simplify.

$v(-u + 2f) = -u(2f - v)$

$-uv + 2fv = -2fu + uv$

$2fu + 2fv = 2uv$

$fu + fv = uv$

Step 7: Divide both sides by $uvf$.

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

This is the same mirror formula as for a concave mirror. It works for both mirrors as long as the sign convention is used.

Magnification of a convex mirror

From equation (2), the ratio of image height to object height is:

$\frac{A'B'}{AB} = \frac{PB'}{PB}$

With $A'B' = h'$ (positive, erect image), $AB = h$, $PB' = v$ and $PB = -u$:

$m = \frac{h'}{h} = \frac{v}{-u} = -\frac{v}{u}$

For a convex mirror, $u$ is negative and $v$ is positive, so $-\frac{v}{u}$ is positive. The image is also always smaller than the object, so:

$0 < m < 1$

A positive $m$ means the image is virtual and erect; $m$ less than 1 means it is diminished.

Check with numbers: A convex mirror has $f = +15$ cm and an object is at $u = -10$ cm. Then $\frac{1}{v} = \frac{1}{15} - \frac{1}{-10} = \frac{1}{15} + \frac{1}{10} = \frac{1}{6}$, so $v = +6$ cm (behind the mirror). Magnification $m = -\frac{6}{-10} = +0.6$: erect and diminished. ✓

Summary table

Exam tips

• For numericals, always write the sign of every quantity before you substitute into $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
• For a convex mirror, $f$ is positive. Forgetting this is the most common source of wrong answers.
• If you get a negative $v$ for a convex mirror with a real object, recheck your signs: the image must be behind the mirror.
• If asked to derive, name the similarity criterion (AA) and state which angles are equal.

Common mistakes

• Taking $f$ as negative for a convex mirror (that is the sign for a concave mirror).
• Writing $B'C = PC + PB'$. Since $B'$ lies between $P$ and $C$, $B'C = PC - PB'$.
• Forgetting that $m = -\frac{v}{u}$ already contains a minus sign, and adding another one.
• Thinking a convex mirror can form a real image of a real object. It cannot.

Frequently Asked Questions

What is the mirror formula?

The mirror formula is 1/v + 1/u = 1/f, where u is the object distance, v is the image distance and f is the focal length, all measured from the pole with the New Cartesian sign convention.

Is the mirror formula the same for convex and concave mirrors?

Yes. The same formula works for both. Only the signs differ: f is positive for a convex mirror and negative for a concave mirror.

Why is the focal length of a convex mirror positive?

The focus of a convex mirror lies behind the mirror, in the direction of the incident light. Distances in that direction are taken as positive.

Why is the magnification of a convex mirror always positive and less than 1?

A convex mirror always forms an erect, virtual, diminished image of a real object, so h′ is positive and smaller than h. That makes m = h′/h positive and less than 1.

Is the derivation of the mirror formula in the Class 10 syllabus?

No. The rationalised NCERT Class 10 book only states the mirror formula and uses it in problems. The derivation is covered in Class 12 Physics.