the power of the lenses of a certain physicians eyes is 51.5 D. assume the dista
ID: 1793610 • Letter: T
Question
the power of the lenses of a certain physicians eyes is 51.5 D. assume the distance between a retina and a lens in the human eye is 2.00 cm. When she carefully examines a patient, how far, in meters, must the patient be from her? the power of the lenses of a certain physicians eyes is 51.5 D. assume the distance between a retina and a lens in the human eye is 2.00 cm. When she carefully examines a patient, how far, in meters, must the patient be from her? the power of the lenses of a certain physicians eyes is 51.5 D. assume the distance between a retina and a lens in the human eye is 2.00 cm. When she carefully examines a patient, how far, in meters, must the patient be from her?Explanation / Answer
Given, power of the eye lens = 51.5 D
So, focal length = 1/P = (1/51.5) m = 0.0194 m=1.94cm
Ratina distance or image distance = 2 cm
Now, by using lens formula, 1/v - 1/u = 1/f
=> 1/2 - 1/u = 1/1.94
=> 1/v = 1/2 + 1/1.94
=> 1/v = 1.194 + 2/ 2 ×1.194
=> 1/v = 3.194/ 2.388
=> v = 0.747 = 0.75
So patient distance from her eye = 0.75 cm
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