Final answer:
To correct the physicist's myopic eye, a lens with a specific focal length and power is required. By calculating the image distance considering the far point and the distance of the lens from the eye, the focal length, which will be negative, is determined, indicating a diverging lens. The power is then found using the inverse of the focal length in meters.
Step-by-step explanation:
The question asks about determining the focal length and power of a corrective lens that can help a nearsighted (myopic) physicist see clearly at a distance beyond their far point. Given that the far point is 21 cm and the lens will be 2 cm from the eye, we use the formula 1/f = 1/do + 1/di, where do is the distance to the object (in this case, the far point of the eye), and di is the image distance which should be the distance from the lens to the eye.
To correct myopia, the lens must form an image at the far point of the physicist's eye when looking at infinity. Since the far point is the maximum distance at which the eye can see clear images, the image distance di will be equal to the far point distance minus the distance of the lens from the eye, resulting in di = 21 cm - 2 cm = 19 cm. Therefore, we calculate the focal length using the formula with di as 19 cm and do approaching infinity (as do is very large for objects at infinity). The equation simplifies to 1/f ≈ -1/di.
After calculation, the focal length f is negative showing that a diverging lens is needed. Finally, to find the power P of the lens, we use the formula P = 1/f (with f in meters to get power in diopters). With all these steps, the power of the lens is determined.