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CBSE Class 10 Answered

a person can see cannot se object beyond 1.2 m distinctly . what kind of lens should be used to correct his vision and find the focal lenght of lens.
Asked by Hritik Bhardwaj | 09 Mar, 2015, 06:06: PM
answered-by-expert Expert Answer
begin mathsize 14px style The space person space is space suffering space from space myopia space left parenthesis short minus sightedness right parenthesis.
For space normal space eye comma
straight u equals negative infinity
Given colon space straight v equals negative 1.2 space straight m
Now comma
1 over straight f equals 1 over straight v minus 1 over straight u
space space space space space equals fraction numerator 1 over denominator negative 1.2 end fraction minus fraction numerator 1 over denominator negative infinity end fraction
space space space space space equals negative fraction numerator 1 over denominator 1.2 end fraction plus 0
space space space space space equals negative fraction numerator 1 over denominator 1.2 end fraction
therefore The space focal space length space of space the space required space len s comma space straight f equals negative 1.2 space straight m.
So comma space straight a space concave space lens space of space focal space length space 1.2 space straight m space is space to space be space used. end style
Answered by Faiza Lambe | 09 Mar, 2015, 10:50: PM
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