1.

Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.

Answer»

SOLUTION :The diagram is as shown below

`x=1m=100cm`
`d=25cm`
`f=(xd)/(x-d)=(100xx25)/(100-25)`
`f=33.33cm`
`P=(1)/(f(m))`
`=(100)/(f(CM))=(100)/(33.33)=+3D`


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