An object is placed at a distance of 10 cm from a convex lens. The lens produces a real and magnified image with a magnification \(-3\). What is the position of the image?
\(+30.0\text{ cm}\)
For a lens, magnification is \(m = \frac{v}{u}\), where \(u\) is the object distance and \(v\) is the image distance. By the sign convention, a real object placed on the left of the lens has a negative object distance, so \(u = -10\text{ cm}\).
Substituting \(m = -3\): \(v = m \times u = (-3) \times (-10) = +30\text{ cm}\).
The positive sign shows that the image is formed 30 cm on the other side of the lens, which is where a real image is formed by a convex lens, and the negative magnification confirms it is inverted and real.
Check: from the lens formula \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{30} + \frac{1}{10} = \frac{4}{30}\), so \(f = 7.5\text{ cm}\), a positive focal length as expected for a convex lens. The object at 10 cm lies between \(f\) and \(2f\) (7.5 cm and 15 cm), which indeed gives a real, magnified image.
Hence, \(+30.0\text{ cm}\).
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