Shackleford
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21. Two 25cm focal length converging lenses are placed 16.5cm apart. An object is placed 35cm in front of one lens. Where will the final image formed by the second lens be located? What is the total magnification?
HERE IS WHAT I NEED CHECKED. Because the image formed by the first lens is on the opposite side of the second lens from where the light is emanating (right side of second lens), I give the new object distance for the second lens a negative sign. Plugging those numbers into the thin lens equation, I get an image distance of .1848 m. This has a positive sign because it is on the opposite side from where the light is originally emanating.
HERE IS WHAT I NEED CHECKED. Because the image formed by the first lens is on the opposite side of the second lens from where the light is emanating (right side of second lens), I give the new object distance for the second lens a negative sign. Plugging those numbers into the thin lens equation, I get an image distance of .1848 m. This has a positive sign because it is on the opposite side from where the light is originally emanating.