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CBSE Class 10 Answered

 A person is suffering from myopia as well as hypermetropia. He needs lenses of power -5.5D and +1.5D for  correcting myopia and hypermetropia respective. Find the focal length of the  lenses used by him in spectacle to correct myopia and hypermetropia. 
 
Asked by sangeeta | 04 Mar, 2016, 01:11: PM
answered-by-expert Expert Answer
The power of a lens is given as reciprocal of focal length in metres.
 
begin mathsize 14px style straight P equals fraction numerator 1 over denominator straight f left parenthesis straight m right parenthesis end fraction therefore straight f equals 1 over straight P Therefore comma space the space focal space length space of space lens space used space to space correct space myopia space is straight f subscript straight m equals fraction numerator 1 over denominator negative 5.5 end fraction equals 0.182 space straight m equals 18.2 space cm Similarly comma space the space focal space length space of space lens space used space to space correct space hypermetropia space is straight f subscript straight h equals fraction numerator 1 over denominator 1.5 end fraction equals 0.667 space straight m equals 66.7 space cm end style
Answered by Romal Bhansali | 04 Mar, 2016, 06:20: PM

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