Light – Reflection and Refraction • Topic 2 of 3

Mirror Formula & Magnification

Drawing ray diagrams tells us the nature of an image, but to find exact distances we need numbers and a consistent sign system. The New Cartesian Sign Convention fixes the rules. The pole of the mirror is taken as the origin and the principal axis as the x-axis. All distances are measured from the pole. Distances measured against the direction of incident light (i.e. to the left, behind the mirror in the usual diagram) are negative, while distances measured along the incident light are positive. Heights measured upwards (above the axis) are positive and those measured downwards are negative.

Because the object is always placed in front of a mirror, the object distance $u$ is always negative. A concave mirror has a negative focal length; a convex mirror has a positive focal length. These signs are not optional — putting in the wrong sign is the single most common mistake in numericals.

The mirror formula links the object distance, image distance and focal length:

  • $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$, where $v$ is the image distance, $u$ is the object distance and $f$ is the focal length.

The magnification (m) tells us how large the image is compared with the object, and whether it is erect or inverted. It is defined as the ratio of image height $h'$ to object height $h$, and is also related to the distances:

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

The sign of $m$ carries meaning. A negative magnification means the image is real and inverted; a positive magnification means the image is virtual and erect. If $|m| > 1$ the image is enlarged, if $|m| < 1$ it is diminished, and if $|m| = 1$ it is the same size as the object. To solve a numerical, first write down every known quantity with its correct sign, substitute into the mirror formula to find the unknown distance, and then use the magnification relation. Always interpret the final signs to describe the image fully — never report a bare number without saying whether the image is real or virtual, erect or inverted.

Sign convention: object distance u and image distance v measured from the poleP (origin)Object (h)Image (h')u (negative)v (negative)+ direction of incident light →
1
Worked Example
An object is placed $20\,\text{cm}$ in front of a concave mirror of focal length $15\,\text{cm}$. Find the position of the image.
Solution
  1. Sign convention: $u = -20\,\text{cm}$, $f = -15\,\text{cm}$ (concave).
  2. Mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$, so $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-15} - \frac{1}{-20}$.
  3. $\frac{1}{v} = -\frac{1}{15} + \frac{1}{20} = \frac{-4 + 3}{60} = -\frac{1}{60}$.
  4. $v = -60\,\text{cm}$.

Answer: $v = -60\,\text{cm}$ — the image is real, formed $60\,\text{cm}$ in front of the mirror.

2
Worked Example
For the mirror in Example 1 ($u = -20\,\text{cm}$, $v = -60\,\text{cm}$), find the magnification and describe the image.
Solution
  1. Magnification $m = -\frac{v}{u} = -\frac{-60}{-20}$.
  2. $m = -\frac{60}{20} = -3$.
  3. The magnification is negative, so the image is real and inverted; $|m| = 3 > 1$, so it is enlarged three times.

Answer: $m = -3$; the image is real, inverted and three times the size of the object.

3
Worked Example
An object $5\,\text{cm}$ tall is placed $25\,\text{cm}$ in front of a convex mirror of focal length $20\,\text{cm}$. Find the image distance.
Solution
  1. Sign convention: $u = -25\,\text{cm}$, $f = +20\,\text{cm}$ (convex).
  2. $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{20} - \frac{1}{-25} = \frac{1}{20} + \frac{1}{25}$.
  3. $\frac{1}{v} = \frac{5 + 4}{100} = \frac{9}{100}$.
  4. $v = \frac{100}{9} \approx +11.1\,\text{cm}$.

Answer: $v \approx +11.1\,\text{cm}$ — the image is virtual, behind the mirror.

4
Worked Example
For the convex mirror in Example 3 ($u = -25\,\text{cm}$, $v = +11.1\,\text{cm}$, $h = 5\,\text{cm}$), find the magnification and the height of the image.
Solution
  1. $m = -\frac{v}{u} = -\frac{11.1}{-25} = +0.44$.
  2. The magnification is positive, so the image is virtual and erect; $|m| < 1$, so it is diminished.
  3. Image height $h' = m \times h = 0.44 \times 5 = 2.2\,\text{cm}$.

Answer: $m = +0.44$; the image is virtual, erect and $2.2\,\text{cm}$ tall.

5
Worked Example
An object is placed $10\,\text{cm}$ in front of a concave mirror of focal length $15\,\text{cm}$. Find the image distance and nature.
Solution
  1. Sign convention: $u = -10\,\text{cm}$, $f = -15\,\text{cm}$.
  2. $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-15} - \frac{1}{-10} = -\frac{1}{15} + \frac{1}{10}$.
  3. $\frac{1}{v} = \frac{-2 + 3}{30} = \frac{1}{30}$, so $v = +30\,\text{cm}$.
  4. $v$ is positive (behind the mirror), so the image is virtual.

Answer: $v = +30\,\text{cm}$ — the image is virtual and erect (object lies between P and F).

6
Worked Example
A concave mirror produces a real image magnified two times of an object placed $30\,\text{cm}$ in front of it. Find the focal length.
Solution
  1. Real, inverted image means $m = -2$. Given $u = -30\,\text{cm}$.
  2. $m = -\frac{v}{u}$, so $-2 = -\frac{v}{-30}$, giving $v = -60\,\text{cm}$.
  3. $\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{-60} + \frac{1}{-30} = -\frac{1}{60} - \frac{2}{60} = -\frac{3}{60} = -\frac{1}{20}$.
  4. $f = -20\,\text{cm}$.

Answer: Focal length $= -20\,\text{cm}$ (concave mirror).

Key Points

  • New Cartesian sign convention: pole is the origin; distances along the incident light are positive, against it negative; heights up are positive, down negative.
  • Object distance $u$ is always negative; concave mirror $f$ is negative, convex mirror $f$ is positive.
  • Mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
  • Magnification: $m = \frac{h'}{h} = -\frac{v}{u}$.
  • Negative $m$ → real and inverted image; positive $m$ → virtual and erect image.
Tap an option to check your answer0 / 4
Q1.In the New Cartesian sign convention, the object distance for a mirror is always:
Explanation: The object is placed in front of the mirror, against the incident light, so $u$ is always negative.
Q2.The mirror formula is:
Explanation: For mirrors the formula is $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Q3.A magnification of $-2$ means the image is:
Explanation: Negative $m$ means real and inverted; $|m| = 2 > 1$ means enlarged.
Q4.An object is $20\,\text{cm}$ from a concave mirror of focal length $10\,\text{cm}$ ($u=-20$, $f=-10$). The image distance $v$ is:
Explanation: $\frac{1}{v} = \frac{1}{-10} - \frac{1}{-20} = -\frac{1}{20}$, so $v = -20\,\text{cm}$ (object at C gives image at C).