Explanation
Hypermetropia is a condition of the eye where the person is not able to see things clearly when nearer to the eye.
The normal near point of the eye is =$$25cm$$
For corrective lens to be used ,
Focal length,= $$f$$
For corrective lens object distance =The normal near point of the eye = $$u=-25cm$$ and image distance = defected near point of man = $$ v=-100cm$$
Lens formula:
$$ \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u} $$
$$ \dfrac{1}{f}=\dfrac{1}{-100}-\dfrac{1}{-25}=\dfrac{1}{25}-\dfrac{1}{100}=\dfrac{3}{100} $$
$$ f=33.33\,cm=0.33\,m $$
Power, $$P=\dfrac{1}{f}=\,\dfrac{1}{0.333}=3D$$
Hence, Power of lens is $$3D$$
Given,
Near point, $$v=-120\,cm$$
Reading Distance, $$u=-40\,cm$$
From lens formula,
$$ \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{-120}-\dfrac{1}{-40}=\dfrac{1}{60} $$
$$ f=+60\,cm $$
Focal length of lens $$f=+60\,cm$$
From the question we can see that the student is unable to see objects that are away from him clearly. This is the condition for myopia.
Hence the correct answer is option (A).
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