# How to solve numericals related to myopia and hypermetropia? Plz explain.

Hope you got it.

@Siddu, We appreciate your contribution to this forum.

• -29

Well I guess there will be no prob with numericals in case of Eye Defects

We are Asked to state the type of lens which is used and  Draw the Diagram for the defected  eye and its correction ..

• 7

For Myopic eye

Power of the correcting concave lens

The lens formula can be used to calculate the focal length and hence the power of the myopia correcting lens.

In this case,

Object distance, u =

Image distance, v = person’s far point

Focal length, f =?

Hence, lens formula becomes

In case of a concave lens, the image is formed in front of the lens i.e., on the same side of the object.

∴

Now, Power of the required lens (P) =

Example: A person can clearly see up to a maximum distance of 100 cm only. Calculate the power of the required lens that can correct his defect?

Solution:

Since the person is not able to see farther than 100 cm, he is suffering from myopia. Hence, a concave lens of suitable power is required to correct his defect. The focal length of the lens is given by his far point i.e.,

Focal length =  Far point

100 cm

∴ Power of the lens =

Hence, a concave lens of power 1 D is required to correct the given defect of vision.

• 91

For Hypermetropia

Power of the correcting convex lens

Lens formula,  can be used to calculate focal length and hence power of the correcting convex lens, where

Object distance, u = –25 cm, normal near point

Image distance, v = defective near point

Hence, the lens formula is reduced to

Example: The defective near point of an eye is 150 cm. Calculate the power of the correcting convex lens that would correct this defect of vision.

Solution:

Given that, hypermetropic near point = 150 cm

Hence, image distance, v = – 150 cm

We have the correction formula,

∴ Power of the correcting convex lens,

P =

Hence, a convex lens of power 3.3 D is required to correct the given defect of vision.

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