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Question: A hypermetropic person has to use a lens of power +3D to normalize his vision. The near point of the...

A hypermetropic person has to use a lens of power +3D to normalize his vision. The near point of the hypermetropic eye is:
A: 1m
B: 1.5m
C: 0.5m
D: 0.66m

Explanation

Solution

Hypermetropia is commonly referred to as long sightedness or far sightedness. It is a condition when the nearby objects appear unclear but the vision is clearer as the object is farther. This condition is corrected using a convex lens where the light rays are focussed to the retina.
Formula used:
Lens formula: 1f=1v1u\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}
Where f is the focal length of the lens, v is the image distance and u is the object distance.

Complete step by step answer:
We know that the normal near point of the eye is 25cm, that is the object must be at a distance of at least 25cm.
We are given the power of the lens, which is +3D.
P=1f f=13m=1003cm \begin{aligned} & P=\dfrac{1}{f} \\\ & \Rightarrow f=\dfrac{1}{3}m=\dfrac{100}{3}cm \\\ \end{aligned}
When we consider the object distance u to be 25cm, i.e. the near point, we can find the image distance v from the lens formula, that is
1f=1v1u\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

Conventionally, the object distance is taken as negative.
Hence,
1100/3=1v125 3100=1v+125 1v=3100125=1100  \dfrac{1}{100/3}=\dfrac{1}{v}-\dfrac{1}{-25} \\\ \Rightarrow \dfrac{3}{100}=\dfrac{1}{v}+\dfrac{1}{25} \\\ \Rightarrow \dfrac{1}{v}=\dfrac{3}{100}-\dfrac{1}{25}=\dfrac{-1}{100} \\\
Hence the near point of the hypermetropic eye is at 100cm or 1m from the eye.

So, the correct answer is “Option A”.

Note:
Convex lenses are used to treat hypermetropia as it helps to focus the light rays on the retina. It is also used in cameras to help in focussing the light for a clear image. It is also used in compound lenses that are used in magnifying devices like microscopes and telescopes.