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The graph shown relationship between object distance and image distance for a equiconvex lens. Then focal length of the lens is ` A. `0.50+-0.05cm`B. `0.50+-0.10cm`C. `5.00+-0.05cm`D. `5.00+-0.10cm` |
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Answer» Correct Answer - c. From the formula `(1)/(f)=(1)/(v)-(1)/(u)`, we have `(1)/(f)=(1)/(10)-(1)/(-10)rArrf=+5` Further, `Deltau=0.1` and `Deltav=0.1` (from the graph) Now, differentionting the lens formula, we have `(Deltaf)/(f^(2))=(Deltav)/(v^(2))+(Deltau)/(u^(2))` `Deltaf+((Deltav)/(v^(2))+(Deltau)/(u^(2)))f^(2)` Substituting the values, we have `Deltaf=((0.1)/(10^(2))+(0.1)/(10^(2)))(5)^(2)=0.05` `:.f+-Deltaf=5+- =0.5` |
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