Question

a thin converging lens is placed 20.0 cm from a vertical object and a clear, real image of that object appears on a screen 100.0 cm behind the lens. the positions of the object and screen are fixed. which new location can the lens be moved to so that a clear real image appears on the screen again (e.g. you can specify whether the lens should be moved further or closer to the screen, and the distance)? what are the transverse magnifications of the two images?

Answer 1

To solve for the new location of the lens and the transverse magnifications, the lens equation and magnification formula must be applied twice, using the given distances of the object and the screen.

Step-by-step explanation:

The student's question pertains to a thin converging lens and involves finding new positions for the lens where a real image is formed, as well as calculating the transverse magnifications of the images. Given the fixed positions of the object and screen, we can use the lens equation 1/f = 1/do + 1/di to find the new lens location. The magnification m can be found using the equation m = -di/do. To solve the problem, we need to find two solutions to the lens equation where di is 100 cm and thus determine the corresponding object distances do. With these, we can calculate the respective magnifications.

For the first case, where di is 100 cm and do is 20 cm, the lens equation gives us f, the focal length of the lens. Using the focal length derived, we can find the second position of the lens that also produces an image at di = 100 cm with the second solution to the lens equation. Lastly, we plug each value of di and the corresponding do into the magnification equation to find the magnifications of the two images.