Formula
The reciprocals of the focal, object and image distances of a thin lens are linked by: 1/F = 1/d + 1/f.
ENT-level problem
An object is placed at a distance $d = 30$ cm from a converging lens of focal length $f = 10$ cm. Find the distance $d'$ from the lens to the image of the object.
The thin lens formula:
$$\frac{1}{f} = \frac{1}{d} + \frac{1}{d\prime}$$
$$\frac{1}{d\prime} = \frac{1}{f} - \frac{1}{d} = \frac{1}{10} - \frac{1}{30} = \frac{3 - 1}{30} = \frac{2}{30} = \frac{1}{15}$$
$$d\prime = 15 \text{ cm}$$
The thin lens formula links the focal length F with the object distance d and the image distance f. All three lengths are written in the same units — centimetres are the most convenient for ENT tasks.
If the object is beyond the focus (d > F), the image is real and inverted; if it is closer (d < F), the image is virtual and upright (this is how a magnifying glass works). At d = 2F the image is the same size as the object.
Solve the formula for F: F = d · f / (d + f). For d = 30 cm and f = 15 cm the focal length is 10 cm.
When the object is inside the focus (d < F): the image is virtual, upright and magnified — this is how a magnifying glass works.
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