The distance between an object and ts upright image is 20.0 cm. If the magnification is 0.500, what is the magnitude of the focal length of the lens that is being used to form the image?

The lens being used in this scenario is likely a converging lens, as a diverging lens would not result in an upright image. Additionally, since the image is diminished, it must be located farther from the lens than the object. Using the thin lens equation, 1/f = 1/s + 1/s', where f is the focal length, s is the distance from the lens to the object, and s' is the distance from the lens to the image. We are given that s' - s = 20 cm and m = -s'/s = 0.500. Since the image is upright, we use a negative sign in front of the magnification. Rearranging the equation for magnification, we get s' = -0.500s. Substituting this into the equation for the distance between the object and image, we get: s' - s = -1.500s = 20 cm Solving for s, we get s = -20/1.5 = -13.33 cm (the negative sign indicates that the object is located to the left of the lens) Now we can use the thin lens equation to find the focal length: 1/f = 1/s + 1/s' = 1/-13.33 + 1/(-0.500*-13.33) = -0.076 + 0.150 Therefore, f = 6.67 cm.