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Given refractive index is 3/2 and refractive index of water =4/3.
if a convex lens of focal length 10 centimetre is placed in water then its focal length in water is???
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If n is refractive index then
1/f=(n-1)(1/R’-1/R”)
let 1/R’-1/R”)=k.
then 1/fa=[n(ag) -1] k
if n(ag) = 3/2, n(Aw) =4/3
then n(wg) =n(ag) /n(aw)
=(3/2)*(3/4)=9/8
if fa=10cm
here fa is focal length of glass lens in air and
fw is focal length of glass lens in water then
1/fw=[n(wg)-1]k
1/fw=(9/8-1)k and
1/10=(3/2-1)k
dividing
10/fw=(9/8-1)/(3/2-1)
=(1/8)/(1/2)
10/fw=1/4
fw=10*4
fw=40cm. Answer
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roshan030:
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