InterviewSolution
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InterviewSolution
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Calculate the power of the eye lens of the normal eye, when it is focused at far point and near point, given the diameter of the eye is ` 2.5 ` cm. Find the maximum variation in the power of normal eye lens. |
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Answer» The far point of a normal eye is infinity. When the object is at infinity, the image is formed at the focus, i.e., image distance v = f, where f is the focal length. Diameter of the eye = distance between lens and the focus = ` 2.5` cm(given) ` :. f = 2.5 cm = 2/5 xx 10^(-2)` m . Power of the lens, ` p= 1/f` ` p = 1/(2.5 xx 10^(-2)) = 10^(2)/(2.5) = 40` D. 2. The near point of a normal eye is 25 cm = object distance = - 25 cm = u (form sign convention) ` v = 2.5 ` cm = distance of the eye lens from the retina 9 i.e., the focus) ` 1/f = 1/(2.5 xx 10^(-2)) - 1/(-25 xx 10^(-2))` ` 1/f = 100/2.5 + 100/25` ` 1/f = 40 + 4` ` 1/f = 44` Power = ` 1/f = 44` D. Thus, the maximum variation in the power of the lens is ` 44 D - 40 D = 4 D`. |
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