Final answer:
To find the final image distance from the converging lens, the thin-lens equation is used for both the diverging and converging lenses in succession. The distance of the first image from the diverging lens becomes the object distance for the converging lens after adjusting for the lens separation. The calculated distance from the converging lens to the image is positive for real images and negative for virtual ones.
Step-by-step explanation:
To determine how far to the right of the converging lens the final image of a flower will be, we start by using the thin-lens equation 1/f = 1/do + 1/di for each lens individually. For a diverging lens, the focal length (f) is taken as negative, while for a converging lens, it's positive. The distance of the object (do) is the distance between the object and the diverging lens, and the distance of the image (di) will be the result we are solving for. The magnification (m) can also be calculated from m = -di/do for each lens separately.
However, this problem requires two steps since there are two lenses. After finding the image distance from the diverging lens, this image serves as a virtual object for the converging lens. The distance between the two lenses must be subtracted from this image distance to find the correct object distance for the second lens. Once we have this new object distance (do') for the converging lens, we use its focal length to find the final image distance (di') relative to the converging lens using the thin-lens equation again.
The sign of the resulting di' tells us the nature of the image. If di' is positive, the image is real and on the opposite side of the light entry, if negative, it's virtual and on the same side as the light entry. The final distance from the converging lens to the real image is the absolute value of di' if di' is positive.