A converging lens has do = 0.30 m and forms an image at di = 0.15 m. What is the focal length f?

A converging lens has do = 0.30 m and forms an image at di = 0.15 m. What is the focal length f?

• A. 0.10 m
• B. 0.20 m
• C. 0.15 m
• D. 0.05 m

Answer: (A)

The distances of the object and its image with a converging lens are related by the thin lens equation: 1/f = 1/do + 1/di. Here, do is 0.30 m and di is 0.15 m. Plugging in gives 1/f = 1/0.30 + 1/0.15 = 3.333… + 6.666… = 10. Therefore f = 0.10 m. This result is consistent with how a converging lens behaves: the focal length must be positive, and with f = 0.10 m, 2f = 0.20 m. Since the object is located at 0.30 m (beyond 2f), the image should form between f and 2f, which is 0.10 to 0.20 m, and the given image distance of 0.15 m lies in that range.

The distances of the object and its image with a converging lens are related by the thin lens equation: 1/f = 1/do + 1/di. Here, do is 0.30 m and di is 0.15 m. Plugging in gives 1/f = 1/0.30 + 1/0.15 = 3.333… + 6.666… = 10. Therefore f = 0.10 m.

This result is consistent with how a converging lens behaves: the focal length must be positive, and with f = 0.10 m, 2f = 0.20 m. Since the object is located at 0.30 m (beyond 2f), the image should form between f and 2f, which is 0.10 to 0.20 m, and the given image distance of 0.15 m lies in that range.