
The near point of a hypermetropic eye is 1m. What is the nature and power of the lens required to correct this defect? (Assume that the near point of the normal eye is 25 cm).
Answer
539.7k+ views
Hint: To solve this problem first we will discuss Hypermetropia and its effect on forming image behind the retina and by using lens formula to calculate power of the lens to correct this defect, we will also discuss about different corrective measures for the same.
Formula used:
Len’s formula,
$\Rightarrow \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$
Power,
$\Rightarrow P=\dfrac{1}{f}$
Complete answer:
Hypermetropia can also be termed as farsightedness in which light rays focus on behind the retina instead of retina which results in blurry image. This is a type of defect not disease so it can be corrected artificially by different types of measures.
As, we know that our eye behaves as convex lens so virtual and erect image will be formed so,
Object distance U=-25 cm,
Image distance V=-1 m = -100 cm
Now, by using lens formula
$\Rightarrow \dfrac{1}{f}=\dfrac{1}{V}-\dfrac{1}{U}$
$\Rightarrow \dfrac{1}{f}=\dfrac{1}{(-100)}-\dfrac{1}{(-25)}$
$\Rightarrow \dfrac{1}{f}=-\dfrac{1}{100}+\dfrac{1}{25}$
$\Rightarrow \dfrac{1}{f}=\dfrac{-1+4}{100}=\dfrac{3}{100}$
So, focal length
$\Rightarrow f=\dfrac{100}{3}=33cm=0.33m$
Now, power of the lens is given by,$P=\dfrac{1}{f}$
$\Rightarrow P=\dfrac{1}{0.33}=3D$(Diopters)
So, to correct the defect of this hypermetropic eye lens of 3 diopters will be used.
Note:
Hypermetropic eyes can be corrected by three corrective measures. First, by using corrective eyeglasses, which depends on how much focal length has been changed due to defect, so the convex lenses in eyeglasses can correct this problem and it is the most widely accepted corrective measure. Second, by using contact lenses which provide the same effect as eyeglasses but in a more convenient and comfortable option but little bit expensive as well and for the last option refractive surgery is done to alter the defect of the eye by means of surgery.
Formula used:
Len’s formula,
$\Rightarrow \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$
Power,
$\Rightarrow P=\dfrac{1}{f}$
Complete answer:
Hypermetropia can also be termed as farsightedness in which light rays focus on behind the retina instead of retina which results in blurry image. This is a type of defect not disease so it can be corrected artificially by different types of measures.
As, we know that our eye behaves as convex lens so virtual and erect image will be formed so,
Object distance U=-25 cm,
Image distance V=-1 m = -100 cm
Now, by using lens formula
$\Rightarrow \dfrac{1}{f}=\dfrac{1}{V}-\dfrac{1}{U}$
$\Rightarrow \dfrac{1}{f}=\dfrac{1}{(-100)}-\dfrac{1}{(-25)}$
$\Rightarrow \dfrac{1}{f}=-\dfrac{1}{100}+\dfrac{1}{25}$
$\Rightarrow \dfrac{1}{f}=\dfrac{-1+4}{100}=\dfrac{3}{100}$
So, focal length
$\Rightarrow f=\dfrac{100}{3}=33cm=0.33m$
Now, power of the lens is given by,$P=\dfrac{1}{f}$
$\Rightarrow P=\dfrac{1}{0.33}=3D$(Diopters)
So, to correct the defect of this hypermetropic eye lens of 3 diopters will be used.
Note:
Hypermetropic eyes can be corrected by three corrective measures. First, by using corrective eyeglasses, which depends on how much focal length has been changed due to defect, so the convex lenses in eyeglasses can correct this problem and it is the most widely accepted corrective measure. Second, by using contact lenses which provide the same effect as eyeglasses but in a more convenient and comfortable option but little bit expensive as well and for the last option refractive surgery is done to alter the defect of the eye by means of surgery.
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