The graph shown relationship between object distance and image distance for a equiconvex lens. Then focal length of the lens is `

Options

A. `0.50+-0.05cm`
B. `0.50+-0.10cm`
C. `5.00+-0.05cm`
D. `5.00+-0.10cm`

Answer

c.

Explanation

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`