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+-00.5cm`
B. `0.50+-0.10 cm`
C. `5.00+-0.05 cm`
D. `5.00+-0.10cm^(u)`

Answer

C

Explanation

We know that in case of a convex lens when object is placed at C`, the image is obtained at C. This situation is represented in the graph by the point corresponding to `u=-10cm` , `v=10 cm`.
Therefore `R=10 cm` `implies R/(2)=5 cm=f`
lens formula is
`(1)/(f)=(1)/(v)-(1)/(u)implies (trianglef)/(f^2)=(trianglev)/(v^2)+(triangleu)/(u^2)`
(for maximum error if f)
`implies `(trianglef)/(25)=(1.0)/((10)^2)+(0.1)/((10)^2)` [`triangleu=trianglev=0.1 from graph)`
`implies `trianglef=25xx0.1xx2xx0.01=0.05`
Therefore, the focal length`=(5.00+-0.05)`cm.