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The lens equation and magnification

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The lens equation

$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$

$f$ — focal length of the lens (m or cm)
$d_o$ — object distance, from the object to the lens
$d_i$ — image distance, from the lens to the image
Keep all three in the same unit — all centimetres or all metres. Mixing them is the most common error on this page.
The same equation works for curved mirrors, with the same sign rules.

Magnification

$m = \frac{h_i}{h_o} = \frac{d_i}{d_o}$

$m$ — magnification (no units — it is a ratio)
$h_i$ — image height, $h_o$ — object height
$d_i$ — image distance, $d_o$ — object distance
$m > 1$ : the image is larger than the object.
$m < 1$ : the image is smaller.
$m = 1$ : same size (which happens when the object is at $2F$ ).

Signs: real and virtual

The convention used in NCEA Level 2 is real is positive:

Quantity	Positive when	Negative when
$d_o$	object is real (always, at this level)	—
$d_i$	image is real (far side of the lens)	image is virtual (same side as the object)
$f$	lens is converging	lens is diverging

A negative $d_i$ from your calculation is not a mistake — it is the answer telling you the image is virtual, on the same side as the object.
Report the size of a distance with its nature in words: " $d_i = -12$ cm, so the image is virtual, $12$ cm from the lens on the same side as the object".
Take the magnification as a ratio of sizes and state the orientation in words: a real image from a single converging lens is inverted, a virtual one is upright.

The routine

List $f$ , $d_o$ and anything else given, with signs and consistent units.
Substitute into $\dfrac{1}{f} = \dfrac{1}{d_o} + \dfrac{1}{d_i}$ .
Rearrange to $\dfrac{1}{d_i} = \dfrac{1}{f} - \dfrac{1}{d_o}$ .
Invert at the end — this is the step everyone forgets.
Find the magnification with $m = d_i/d_o$ , then the image height with $h_i = m \times h_o$ .
Describe the image in words: real/virtual, upright/inverted, larger/smaller.

Worked ExampleA real image

An object is placed $30$ cm from a converging lens of focal length $10$ cm. The object is $4.0$ cm tall. Find the image distance, the magnification and the image height, and describe the image.

Step 1 — List what you have

$f = +10 \text{ cm}, \qquad d_o = 30 \text{ cm}, \qquad h_o = 4.0 \text{ cm}$

Step 2 — Rearrange and substitute

$\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{10} - \frac{1}{30}$ $\frac{1}{d_i} = 0.1000 - 0.0333 = 0.0667$

Step 3 — Invert

$d_i = \frac{1}{0.0667} = 15 \text{ cm}$

The positive value means the image is real, $15$ cm from the lens on the far side.

Step 4 — Magnification and image height

$m = \frac{d_i}{d_o} = \frac{15}{30} = 0.50$ $h_i = m \times h_o = 0.50 \times 4.0 = 2.0 \text{ cm}$

Step 5 — Describe it

$d_i = 15 \text{ cm}, \quad m = 0.50, \quad h_i = 2.0 \text{ cm}$

The image is real, inverted, and half the size of the object.

Worked ExampleA virtual image (the magnifying glass)

An object $6.0$ mm tall is placed $4.0$ cm from a converging lens of focal length $6.0$ cm. Find the image distance and the image height, and describe the image.

Step 1 — Note where the object sits

$d_o = 4.0$ cm is less than $f = 6.0$ cm, so the object is inside the focal length. Expect a virtual image.

Step 2 — Substitute

$\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} = \frac{1}{6.0} - \frac{1}{4.0}$ $\frac{1}{d_i} = 0.1667 - 0.2500 = -0.0833$

Step 3 — Invert

$d_i = \frac{1}{-0.0833} = -12 \text{ cm}$

The negative sign means the image is virtual, $12$ cm from the lens on the same side as the object.

Step 4 — Magnification and image height

Using the sizes of the distances:

$m = \frac{d_i}{d_o} = \frac{12}{4.0} = 3.0$ $h_i = 3.0 \times 6.0 = 18 \text{ mm}$

Step 5 — Describe it

$d_i = 12 \text{ cm (virtual)}, \quad m = 3.0, \quad h_i = 18 \text{ mm}$

The image is virtual, upright, and three times the size of the object.

Worked ExampleFinding the focal length

A slide projector forms a sharp real image on a screen $3.0$ m from the lens. The slide is $12$ cm from the lens. Find the focal length of the lens and the magnification.

Step 1 — Get the units consistent

$d_i = 3.0 \text{ m} = 300 \text{ cm}, \qquad d_o = 12 \text{ cm}$

Step 2 — Substitute into the lens equation

$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{12} + \frac{1}{300}$ $\frac{1}{f} = 0.08333 + 0.00333 = 0.08667$

Step 3 — Invert

$f = \frac{1}{0.08667} = 11.5 \approx 12 \text{ cm}$

Step 4 — Magnification

$m = \frac{d_i}{d_o} = \frac{300}{12} = 25$

$f = 12 \text{ cm}, \qquad m = 25$

The lens equation and magnification

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