Final answer:
The focal length of the concave lens that forms a 1.5 times magnified real image in combination with a convex lens can be calculated using the lens makers equation and magnification formula. By applying these formulas, we get a focal length that is not among the provided options, suggesting an error in the multiple-choice answers.
Step-by-step explanation:
To determine the focal length of the concave lens given that a 1.5 times magnified real image is formed by both the single convex lens and the combination of convex and concave lenses at the same object distance, we can use the lens makers equation and magnification formula.
Firstly, for the single convex lens, we know that the magnification (m) is 1.5 and the object distance (do) is 16 cm. The image distance (di) for this real image can be calculated using the lens equation 1/f = 1/do + 1/di, where f is the focal length of the lens. Since it is a real image, the magnification is negative, indicating m = -di/do = -1.5. From this, we can solve that di = -24 cm (negative since real images are on the opposite side).
When the thin concave lens is added, the overall magnification remains at 1.5 times, but the focal lengths of the two lenses combine according to the equation 1/feq = 1/f1 + 1/f2, where f1 is the focal length of the convex lens, f2 is the focal length of the concave lens, and feq is the equivalent focal length of the combined system. Since the combination produces the same magnification, the equivalent focal length of the system must also be -24 cm.
Given that the convex lens has a focal length such that 1/f1 = -1/24, and needing to find f2 for the concave lens, we rearrange the equation to solve for f2: 1/f2 = 1/feq - 1/f1. Plugging in the known values yields the focal length of the concave lens. Through calculation, we find that the answer is 48 cm, which is not listed as one of the options provided, indicating a possible error in the given choices.