Refraction and Lenses Legacy Problem #30 Guided Solution

An effective problem solver will pick up on small cues or hints within the statement of a problem that make all the difference in the world between getting the question right and wrong. Here we read of an upright image being produced that is magnified in size by a factor of 4. That indicates to me that the height divided by the whole ratio is equal to 4. That's the magnification. And it's equal to positive 4, not negative, because it's an upright image and upright images have positive image heights. Now this occurs whenever the object is 26.9 centimeters from the lens. The unknown quantity is the focal length. I know that to find the focal length, I'll have to use the lens equation. The lens equation states that 1 over f equals 1 over d object plus 1 over d image. I know the d object, but I do not know the d image. So the crux of my strategy will be to use the first clause of the first sentence in order to find the image distance. To use the idea that the height divided by whole is equal to positive 4 in order to get d image. So what I know about a height divided by whole ratio is that it's equal to the negative of the di divided by do ratio. So I can say that positive 4 is equal to negative di d image divided by the 26.9 centimeters for d object. I can multiply both sides of the equation by 26.9 centimeters and that will give me the image distance. It comes out to be negative 107.6 meters, or centimeters. I expect the negative because it's a virtual image, as all upright images are, and virtual images have negative image distances. Now that I know both di and do, I can substitute into the lens equation. I say 1 over f is equal to 1 divided by 26.9 plus 1 divided by negative 107.6. I evaluate the right side of the equation, it comes out to be 0.027881, and I take the reciprocal of that number and that gives me the focal length. It comes out to be positive 35.9 centimeters. I would have expected positive because positive focal lengths correspond to converging lenses, and the only time you have an upright image that's magnified is produced by a lens as if the lens is a converging lens. Diverging lenses have negative focal lengths and produce reduced in size virtual images.