Question
Two vertical object pins are located on the principal axis of a thin lens at distance of $x_{1} = 9.0\text{ }cm$ and $x_{2} = 18.0\text{ }cm$ from the lens on the two sides of the lens. the image of the two pins formed by the lens are at the same positions (overlap. the focal length and the nature of the lens are
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Solution
Sol. Key concept: Thin-lens formula with real and virtual images on the same physical side.
Choose the common image on the side of the pin 9 cm from the lens, at distance s from the lens. The nearer pin then gives a virtual image, whereas the pin 18 cm away on the opposite side gives a real image at the same location.
$$\frac{1}{f} = \frac{1}{9} - \frac{1}{s},\ \ \ \ \frac{1}{f} = \frac{1}{18} + \frac{1}{s}$$
Adding eliminates the image distance.
$$\frac{2}{f} = \frac{1}{9} + \frac{1}{18} = \frac{1}{6}\ \ \Rightarrow \ \ \ f = 12\ cm$$
Substitution gives s = 36 cm. A positive focal length means a converging (convex) lens. Thus (b).
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