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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. |
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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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