The graph between object distance `u` and image distance `v` for `a` lens is given below. The focal length of the lens is.
(a) `5 pm 0.1`
(b) `5 pm 0.05`
( c) `0.5 pm 0.1`
(d) `0.5 pm 0.05`.
.

Options

A. `5+-0.1`
B. `5+-0.05`
C. `0.5+-0.1`
D. `0.5+-0.05`

Answer

B

Explanation

`(1)/(f)=(1)/(v)-(1)/(u) ……..(1)`
`(1)/(f)=(1)/(10)-(1)/(-10)`
`f=+5`
By differentiate eq. `(1)`
`rArr(-Deltaf)/(f^(2))=(-Deltav)/(v^(2))+(Deltau)/(u^(2))=(1)/(+v^(2))Deltav+(1)/(u^(2))Deltau`
`rArr+(Deltaf)/(5^(2))=(1xx(0.1))/(+10^(2))+(1xx(0.1))/(10^(2))`
`Deltaf=(0.2)/(100)xx25=(0.2)/(4)=0.05`
So, `f=5+-0.05`