A magnifying glass held 5.00 cm from a coin produces a magnification of 3.00. What is the focal length of the lens?

1. The magnification is positive, so the image is upright and virtual, which is the magnifier arrangement. 2. From m = minus di divided by do, the image distance is di = minus m times do. 3. Substituting with units, di = minus 3.00 times 5.00 cm = -15.0 cm, negative as expected for a virtual image. 4. Now use 1/f = 1/do + 1/di. 5. Substituting, 1/f = 1 divided by 5.00 cm plus 1 divided by (-15.0 cm) = 0.2000 per cm minus 0.0667 per cm = 0.1333 per cm. 6. Inverting, f = 7.50 cm, a positive focal length as a converging lens must have. 7. Answering 15.0 cm stops at the image distance and never applies the thin lens equation. 8. Answering 1.67 cm comes from adding the reciprocals without giving the image distance its minus sign. _Source: OpenStax College Physics (CC BY 4.0), Ch 25 "Geometric Optics", section 25.6 Image Formation by Lenses_