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CBSE Class 8 Answered

A convex lens of N =1.5 and focal length 0.2m is immersed in water of R.I.=1.33..calculate the change in power of the lens
Asked by ampoluharitha | 28 Dec, 2018, 08:53: PM
answered-by-expert Expert Answer
Initial Power of lens = 1/0.2 = 5 Diopter
let us get the radius of curvature of the given lens  from lens makers formula : begin mathsize 12px style 1 over f equals open parentheses mu minus 1 close parentheses open parentheses 1 over R subscript 1 plus 1 over R subscript 2 close parentheses................. left parenthesis 1 right parenthesis end style
where f is focal length, μ is refractive index, Rs are radii of curvature.
 
from (1), by substituting values for f and μ, we get,  begin mathsize 12px style 10 over 2 space equals space open parentheses 1.5 minus 1 close parentheses open parentheses 1 over R subscript 1 plus 1 over R subscript 2 close parentheses space space o r space space open parentheses fraction numerator begin display style 1 end style over denominator begin display style R subscript 1 end style end fraction plus fraction numerator begin display style 1 end style over denominator begin display style R subscript 2 end style end fraction close parentheses space equals space 10 end style ....................(2)
when the lens is immersed in liquid, refractive index becomes = 1.5/1.33 = 1.128
 
hence power of lens when it is immersed in liquid, using eqns.(1) and (2)  : (1.128-1)×10 = 1.28 Diopter
 
Change in power = 5 -1.28 = 3.72 Diopter
Answered by Thiyagarajan K | 29 Dec, 2018, 09:13: AM
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