
An eye has a near point of distance\[0.75m\] . What sort of eye lens in spectacles would be needed to reduce the near point distance of the eye to\[0.25m\] . Also calculate the power of the lens required. Is this eye long-sighted or short-sighted?
Answer
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Hint: We know that a normal eye has the near point at \[25cm\] or \[0.25m\] and it is up to an infinite distance. Hence, in this range, if an object is placed the image of it is created at the retina of the eye. If the range of an eye is less or greater than this range, we call it the defective eye. The important sight-based defects are long sight or hypermetropia, short-sighted or myopia, presbyopia, and astigmatism.
The power can be calculated from the formula of power in terms of the near point of an eye and created the lens to cure the defects. The lens that is needed to cure is either convex or concave as per the defects.
Formula used:
The power of the lens for the near point of the defective eye$(d)$ is greater than the near point of a normal eye is,
$P = 4 - \dfrac{{100}}{d}{\text{dioptre}}$
Complete answer:
Given that, An eye has a near point of distance\[0.75m\]
The near point distance of a normal eye to\[0.25m\]
Hence, the near point of the given eye is greater than that of a normal eye.
This type of defect is called long sight or hypermetropia.
To reduce this defect, the glass of the convex lens is needed.
The power of the lens for the near point of the defective eye$(d)$ is greater than the near point of the normal eye is,
$P = 4 - \dfrac{{100}}{d}{\text{dioptre}}$
Given, \[d = 0.75m = 75cm\]
$\therefore P = 4 - \dfrac{{100}}{{75}}$
$ \Rightarrow P = \dfrac{{300 - 100}}{{75}}$
$ \Rightarrow P = \dfrac{{200}}{{75}}$
$ \Rightarrow P = 2.6{\text{dioptre}}$
Hence, the power of the lens required $ \Rightarrow P = 2.6{\text{dioptre}}$
So, the long-sighted person needs the glass of positive power.
Note:
To reduce the short sight defect, the glass of negative power made up of a concave lens is needed. The power of the concave lens is, $P = - \dfrac{{100}}{d}{\text{dioptre}}$.
To reduce presbyopia, a glass of the bifocal lens is used. A bifocal lens is made up of both concave and convex lenses.
To reduce astigmatism, the glass of the toric lens is used.
The power can be calculated from the formula of power in terms of the near point of an eye and created the lens to cure the defects. The lens that is needed to cure is either convex or concave as per the defects.
Formula used:
The power of the lens for the near point of the defective eye$(d)$ is greater than the near point of a normal eye is,
$P = 4 - \dfrac{{100}}{d}{\text{dioptre}}$
Complete answer:
Given that, An eye has a near point of distance\[0.75m\]
The near point distance of a normal eye to\[0.25m\]
Hence, the near point of the given eye is greater than that of a normal eye.
This type of defect is called long sight or hypermetropia.
To reduce this defect, the glass of the convex lens is needed.
The power of the lens for the near point of the defective eye$(d)$ is greater than the near point of the normal eye is,
$P = 4 - \dfrac{{100}}{d}{\text{dioptre}}$
Given, \[d = 0.75m = 75cm\]
$\therefore P = 4 - \dfrac{{100}}{{75}}$
$ \Rightarrow P = \dfrac{{300 - 100}}{{75}}$
$ \Rightarrow P = \dfrac{{200}}{{75}}$
$ \Rightarrow P = 2.6{\text{dioptre}}$
Hence, the power of the lens required $ \Rightarrow P = 2.6{\text{dioptre}}$
So, the long-sighted person needs the glass of positive power.
Note:
To reduce the short sight defect, the glass of negative power made up of a concave lens is needed. The power of the concave lens is, $P = - \dfrac{{100}}{d}{\text{dioptre}}$.
To reduce presbyopia, a glass of the bifocal lens is used. A bifocal lens is made up of both concave and convex lenses.
To reduce astigmatism, the glass of the toric lens is used.
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