1.

For a person with hypermeteropia, the near point has moved to 1.5 m. calculate the focal length of the correction lens in order to make his eyes normal.

Answer»

SOLUTION :GIVEN that , d= 1.5m, D = 25 cm =0.25m (For ANORMAL EYE The focal length of the correction lens is
`f=(d xxD)/(d-D)`
`=(1.5xx0.25)/(1.5-0.25)`
`=(0.375)/(1.25)=0.3m`


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