The near point of a hypermetropic eye is 1 m. Find the power of the lens required to correct this defect. Assume that near point of the normal eye is 25cm.
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Hypermetropia is a defect of vision in which a person is not able to see objects close to the eye clearly. It can be corrected using a suitable convex lens. Basically, it just converges the rays coming towards the retina so that it falls perfectly on the walls of the retina (While in myopia the image is formed in front of the retina). So, the nearest image a person can see is 100cm away from him. But we want him to see an object placed 25 cm away. Therefore we’ll produce the image of the objects placed at 25 cm to his near point (which is 1 m in this case). Now, put “u”=-25 cm , “v”=-100 cm(using appropriate sign conventions). Using the lens formula 1/f=(1/v)-(1/u), we’ll find out the focal length which happens to be 100/3.
Now, Power is (1 m/f) =100/(100/3)=3D(dioptre).
I hope this answer helped.
Now, Power is (1 m/f) =100/(100/3)=3D(dioptre).
I hope this answer helped.
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