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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 9th March 2015, 6:06 PM
Answered by 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 Expert 9th March 2015, 10:50 PM
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