Two lenses, the first with focal length 10.0 cm and the second with focal length −12.5 cm, are separated by 20.0 cm. An object of height 2.40 cm is placed 15.0 cm in front of the first lens. Which of the following characteristics does the final image have?
A) real and upright
B) real and inverted
C) virtual and upright
D) virtual and inverted
E) virtual and diminished in size

Question

Quizplus · Accepted Answer

ETo determine the characteristics of the final image produced by the system of lenses, we first analyze the image formed by the first lens and then consider this image as the object for the second lens.1. For the first lens (convex, +10.0 cm focal length), we use the lens equation 1/f = 1/v - 1/u, where f is the focal length, v is the image distance (positive for real images, negative for virtual images), and u is the object distance (always positive in these conventions). Plugging in the values, we get 1/10 = 1/v - 1/15, solving for v gives v = 30 cm. This means the first lens produces a real image (since v is positive) 30 cm from the lens on the opposite side of the object. The magnification equation, m = -v/u = -30/15 = -2, tells us this image is inverted (since magnification is negative) and magnified (2 times the object size).2. This image now acts as the object for the second lens. Since the lenses are 20.0 cm apart and the image from the first lens is 30.0 cm away, this "object" for the second lens is 10.0 cm behind the second lens, making it a virtual object for the second lens (we treat this distance as -10.0 cm when plugging into the lens equation for the second lens). The second lens is concave with a focal length of -12.5 cm. Using the lens equation again, 1/f = 1/v - 1/u for the second lens gives 1/-12.5 = 1/v - 1/-10. Solving for v gives v = -25 cm, indicating the final image is virtual (since v is negative) and located 25 cm on the same side of the lens as the object.The magnification of the second lens can be calculated as m = -v/u = -(-25)/(-10) = 2.5, indicating the image is upright (since the magnification is positive) relative to the object for the second lens but still inverted relative to the original object because the first image was inverted. The overall magnification of the system is the product of the magnifications of the two lenses, which is -2 * 2.5 = -5, indicating the final image is inverted relative to the original object and magnified.Therefore, the final image is virtual (since it is formed by the diverging lens and the rays do not actually converge) and inverted relative to the original object. It is also diminished in size compared to the intermediate image but magnified relative to the original object. However, the choices provided seem to mix characteristics of the image after the first lens with those after the second lens. Given the options and the analysis, the best fit for the final image characteristics would be "virtual and inverted" because it is virtual (due to the nature of the image formation by the second lens) and inverted relative to the original object. The option "virtual and diminished in size" could be considered correct if we're focusing on the comparison between the intermediate and final images, but based on standard interpretation, the final image is magnified relative to the original object. Thus, the most accurate description given the options and typical physics conventions is "virtual and inverted." However, this explanation reveals a complexity in interpreting the question based on the provided options.

Two Lenses, the First with Focal Length 10