How to solve numericals related to myopia and hypermetropia? Plz explain.
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?
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.
Power of the correcting convex lens
Lens formula, can be used to calculate focal length f and hence power P 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.
Given that, hypermetropic near point = 150 cm
Hence, image distance, v = – 150 cm
We have the correction formula,
∴ Power of the correcting convex lens,
Hence, a convex lens of power 3.3 D is required to correct the given defect of vision.