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Asked by adjacentcalliber10 | 05 May, 2019, 13:01: PM
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we have lens equation,  begin mathsize 12px style 1 over v minus 1 over u equals space 1 over f end style .................(1)
where v = lens-to-image distance, u = -0.4 m and f = +0.3 m
 
hence we get v = 1.2 m from eqn.(1)
 
If we differentiate eqn.(1), we get, begin mathsize 12px style negative 1 over v squared fraction numerator d v over denominator d t end fraction space plus space 1 over u squared fraction numerator d u over denominator d t end fraction space equals space 0 space space space o r space space space fraction numerator d v over denominator d t space end fraction space equals space v squared over u squared fraction numerator d u over denominator d t end fraction end style................(2)
hence fron eqn.(2) , dv/dt = (1.44/0.16)×0.01 = 0.09 m/s
 
magnification = v/u  = 1.2/(-0.4) = -3
Answered by Thiyagarajan K | 05 May, 2019, 14:01: PM

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