(a) Find the power of the lens necessary to correct an eye with a far point of 2
ID: 2119198 • Letter: #
Question
(a) Find the power of the lens necessary to correct an eye with a far point of 24.5 cm(b) Find the power of the lens necessary to correct an eye with a far point of 48.0 cm.
So I know I need to use Power = 1/focal point. The question is on how to get the focal point. We are given the s values of 1/f = 1/s + 1/s' but I don't know how to get the other s value...
Explanation / Answer
For a near sighted person, only near point is needed for lens. The normal distance between the eye and lens will be 2 cm. Near Point = 18 cm ==> Distance from lens = 18 -2 = 16 cm = 0.16 m Since the object and image distance are on the same side, we have to place a negative sign with the image distance, so Distance from lens = -0.16 m Lens Power = (1/f) = -1.85 1/f = 1/d1 + 1/d2 ==> -1.85 = -1/0.18 + (1/dI) => d1 = 0.27 m Thus, new near point = 0.27 + 0.02 = 0.29 m = 29 cm
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.