A person uses a lens of power + 3D to normalise vision. Near point of hypermetropic eye is(A) 1 m (B) 1.66 m(C) 2 m (D) 0.66 m
Question
A person uses a lens of power + 3D to normalise vision. Near point of hypermetropic eye is(A) 1 m (B) 1.66 m(C) 2 m (D) 0.66 m
Solution
The power of a lens is given by the formula:
P = 1/f
where P is the power of the lens and f is the focal length. The power of the lens is given as +3D, which means the lens is converging and the focal length is positive.
We can rearrange the formula to solve for the focal length:
f = 1/P
Substituting the given power into the formula gives:
f = 1/3 = 0.33 m
The near point of a hypermetropic eye is the farthest point at which the eye can focus. This is typically at the focal length of the lens. Therefore, the near point of the hypermetropic eye is 0.33 m.
However, none of the options provided match this calculation. There might be a mistake in the question or the options provided.
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