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A thin converging glass lens made of glass with refractive index 1.5 has a power of +5.0 D. When this lens is immersed in a liquid of refractive index n, it acts as a divergent lens of focal length 100 cm. What must be the value of n? |
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Answer» <P> Solution :`(1)/(F)=[(n_(2))/(n_(a))-1][(1)/(R_(1))-(1)/(R_(2))]``p=(1)/(f)=5andn_(2)=1.5,n_(a)=1,f=20cm` `(1)/(20)=[(1.5)/(1)-1][(1)/(R_(1))-(1)/(R_(2))]` `(1)/(20)=0.5[(1)/(R_(1))-(1)/(R_(2))]` `(1)/(f_(1))=[(n_(2))/(n_(1))-1][(1)/(R_(1))-(1)/(R_(2))]` `p=(1)/(f_(1))=1,(1)/(f_(1))=-100cm,n_(2)=1.5,n_(1)=n` `(1)/(-100)=[(1.5)/(n)-1][(1)/(R_(1))-(1)/(R_(2))]` Ratio of equation `(2) DIV (1)` `(1)/(-5)=(((1.5)/(n)-1))/(0.5)` `-(0.5)/(5) = (1.5)/(n) - 1` `0.9 = (1.5)/(n)` `n = (1.5)/(0.9) = (5)/(3)` |
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