The mirror formula and magnification

From a drawing to a number

A ray diagram shows the idea, but its answer is not exact. A drawing cannot tell you whether the image is at $29\ \text{cm}$ or $31\ \text{cm}$. For this we need a formula. The formula works for every mirror, concave and convex, if we use one sign convention all the time.

The New Cartesian sign convention

We put the pole $P$ of the mirror at the origin. The principal axis is the $x$-axis. The object is always placed on the left of the mirror, so the light travels from left to right.

• All distances are measured from the pole.
• A distance in the direction of the incident light (to the right of the mirror) is positive.
• A distance against the direction of the incident light (to the left of the mirror) is negative.
• A height above the principal axis is positive. A height below it is negative.

What does this mean for the mirrors?

• The object distance $u$ is always negative, because the object is in front of the mirror.
• For a concave mirror, the focus is in front, so $f$ is negative.
• For a convex mirror, the focus is behind, so $f$ is positive.
• A real image is in front of the mirror, so $v$ is negative. A virtual image is behind the mirror, so $v$ is positive.

The mirror formula

Here $u$ is the object distance, $v$ is the image distance and $f$ is the focal length. Put each with its sign. When the unknown is $v$, we write $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}$.

Magnification

The magnification $m$ tells us how big the image is compared with the object. It is the height of the image $h'$ divided by the height of the object $h$:

• If $|m| > 1$, the image is larger than the object. If $|m| < 1$, it is smaller. If $|m| = 1$, it has the same size.
• If $m$ is positive, the image is upright and virtual. If $m$ is negative, the image is inverted and real.

You should find the two branches of the table from lesson 3. Far objects give images near $F$. As the object nears $F$, the image runs off to infinity. Inside $F$, $v$ turns positive: the image is behind the mirror.

Worked example 1: a shaving mirror

A concave mirror has a focal length of $15\ \text{cm}$. An object $2\ \text{cm}$ tall stands $10\ \text{cm}$ in front of it. Where is the image, and how big is it?

Given: $f = -15\ \text{cm}$, $u = -10\ \text{cm}$, $h = 2\ \text{cm}$.

Step 1. We want $v$, so $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = -\dfrac{1}{15\ \text{cm}} + \dfrac{1}{10\ \text{cm}} = \dfrac{-2 + 3}{30\ \text{cm}} = \dfrac{1}{30\ \text{cm}}$. So $v = +30\ \text{cm}$.

Step 2. $m = -\dfrac{v}{u} = -\dfrac{30\ \text{cm}}{-10\ \text{cm}} = +3$.

Step 3. $h' = m \times h = 3 \times 2\ \text{cm} = 6\ \text{cm}$.

The image is $30\ \text{cm}$ behind the mirror ($v$ is positive), virtual, upright ($m$ is positive) and three times larger. The object is between $F$ and $P$, as in a shaving mirror. This agrees with the ray diagram.

Worked example 2: a convex mirror

A convex mirror has $f = +20\ \text{cm}$. An object is $30\ \text{cm}$ in front of it. Find the image.

Step 1. $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{20\ \text{cm}} + \dfrac{1}{30\ \text{cm}} = \dfrac{3 + 2}{60\ \text{cm}} = \dfrac{1}{12\ \text{cm}}$. So $v = +12\ \text{cm}$.

Step 2. $m = -\dfrac{v}{u} = -\dfrac{12\ \text{cm}}{-30\ \text{cm}} = +0.4$.

The image is $12\ \text{cm}$ behind the mirror, virtual, upright and smaller ($0.4$ times the object).

Practise with new numbers

Try it in Python

The program uses the mirror formula and tells us what the image is like.

Common mistakes

• Forgetting the sign of $u$. The object distance is always negative in this convention.
• Giving a concave mirror a positive $f$. A concave mirror has a negative focal length.
• Saying "the image is $-30\ \text{cm}$ from the mirror". The sign shows the side. The distance is $30\ \text{cm}$, and the image is in front of the mirror.

Remember these

• The object is on the left, so $u$ is negative. A concave mirror has a negative $f$, and a convex mirror has a positive $f$.
• Mirror formula: $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$. Magnification: $m = \dfrac{h'}{h} = -\dfrac{v}{u}$.
• A negative $m$ means a real, inverted image. A positive $m$ means a virtual, upright image.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. What are the signs of $u$ and $f$ for an object in front of a concave mirror?

2. An object is $20\ \text{cm}$ from a concave mirror of focal length $10\ \text{cm}$. Where is the image?

3. What does a magnification of $+0.5$ tell you?

4. A convex mirror gives $v = +8\ \text{cm}$ for an object at $u = -24\ \text{cm}$. Find $m$.