A convex lens forms a real, inverted image three times larger than the object. If the object is placed 30 cm from the lens, what is the closest approximate value of the power of the lens?
+4.44 D
Use the Cartesian sign convention with light travelling left to right: object distance \(u = -30\) cm, and for a real inverted image three times larger the linear magnification is \(m = v/u = -3\), giving image distance \(v = -3u = +90\) cm.
Apply the thin-lens formula \(\tfrac{1}{v} - \tfrac{1}{u} = \tfrac{1}{f}\): \(\tfrac{1}{90} - \tfrac{1}{-30} = \tfrac{1}{90} + \tfrac{3}{90} = \tfrac{4}{90}\), so \(f = 22.5\) cm \(= 0.225\) m.
Power of the lens is \(P = 1/f\) with \(f\) in metres, so \(P = 1/0.225 \approx +4.44\) D, positive because the converging lens has a positive focal length.
Hence, the closest approximate value of the power of the lens is +4.44 D.
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