Final answer:
The final image location is approximately 14.92 cm from the converging lens on the opposite side, and the image size is about 2.56 cm.
Step-by-step explanation:
An object of height 3.0 cm is placed at 25 cm in front of a diverging lens with a focal length of 20 cm. To find the location and size of the image created by the first lens (diverging lens), we use the thin-lens equation 1/f = 1/do + 1/di. Calculating for the diverging lens, the thin-lens equation becomes 1/(-20) = 1/25 + 1/di, which gives us the image distance (di) as being -16.67 cm.
The negative sign indicates that the image is virtual and located on the same side as the object. Next, we calculate the magnification (m) using m = -di/do, which gives us a magnification of -16.67/-25 = 0.67. The image height can be found using the magnification, which is 0.67 * 3.0 cm = 2.0 cm.
Assuming that the diverging lens creates a virtual image that acts as an object for the converging lens, we need to consider that the virtual object for the converging lens is its image distance minus the lens separation, which is 16.67 cm - 5 cm = 11.67 cm. The object distance (do) for the converging lens is therefore -11.67 cm (negative because the image is virtual and on the same side as the object).
Using the thin-lens equation for the converging lens, we have 1/20 = -1/11.67 + 1/di, which gives us the final image distance (di) as approximately 14.92 cm from the converging lens on the opposite side. To find the final magnification for the converging lens, we use m = -di/do, and we get -14.92/-11.67 which is approximately 1.28. Therefore, the final image height is 1.28 * 2.0 cm = 2.56 cm.