Final answer:
The object in focus on the retina, given a lens power of 53.0 diopters, is approximately 0.343 meters or 34.3 cm away from the eye. This calculation was done using the relationship between lens power and focal length and the lens formula.
Step-by-step explanation:
The question relates to the optical function of the human eye and involves calculating the distance of an object that is in sharp focus on the retina when the eye's lens power is given. The total effective lens strength or lens power of the eye is stated to be 53.0 diopters (D). Since lens power P in diopters is the inverse of the focal length f in meters (P = 1/f), we can use this relationship to find the focal length of the lens.
f = 1/P
For a lens power of 53.0 D, the focal length f is calculated as:
f = 1/53.0 D = 0.0189 meters (approximately)
Since the object distance (distance from the lens to the object in focus, do) and the image distance (the fixed lens-to-retina distance, which is 2.00 cm or 0.02 meters) are related by the lens formula (1/do + 1/di = 1/f), and the image distance di is equal to the lens-to-retina distance, we can solve for object distance do:
1/do + 1/0.02 m = 1/0.0189 m
1/do = 1/0.0189 m - 1/0.02 m
1/do = 52.91 m-1 - 50 m-1
1/do = 2.91 m-1
do = 1/2.91 m-1 = 0.343 m
So the object being focused on the retina is approximately 0.343 meters, or 34.3 cm, away from the eye.