The near point of a hypermetropic eye is 1m the power of the lens required to correct this defect is

"The near point of a hypermetropic eye is 1 m . what is the power of the lens required to correct this defect?" In this question why the image distance is to be taken as negative,on what basis the negative sign has been chosen?

Suggest Corrections

Answer

Verified

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 cmNow, 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.

Text Solution

Solution : The eye defect called hypermetropia is corrected by using a convex lens. So, the person requires convex lens spectacles. We will first calculate the focal length of the convex lens required in this case. This hypermetropic eye can see the nearby object kept at 25 cm (at near point of normal eye ) clearly if the image of this object is formed at its own near point which is 1 metre here. So, in this case : Object distance, u = -25 cm `" "` (Normal near point) Image distance, v = -1m `" "` (Near point of this defective eye) = - 100 cm And, Focal length, `f = ?" "` (To be calculated) Putting these value in this lens formula, `(1)/(v)-(1)/(u) = (1)/(f)` We get : `(1)/(-100) - (1)/(-25) = (1)/(f)` or `- (1)/(100) + (1)/(25) = (1)/(f)` `(-1 + 4)/(100) = (1)/(f)` `(3)/(100) = (1)/(f)` `f = (100)/(3)` `f = 33.3 cm` Thus, the focal length of the convex lens required is + 33.3 cm. We will now calculate the power. Please note that 33.3 cm is equal to `(33.3)/(100)` m or 0.33 m. Now, Power, `P = (1)/(f ("in metres"))` `= (1)/(+ 0.33)` `= + (100)/(33)` `= + 3.0 D` So, the power of convex lens required is + 3.0 dioptres. Myopia and hypermetropia are the two most common defects of vision (or defects of eye). We will now study another defect of vision which occurs in old age. It is called presbyopia.