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Q.
The graph between object distance $u$ and image distance $v$ for a lens is given below. The focal length of the lens is
IIT JEEIIT JEE 2006
Solution:
From the lens formula,
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ we have,
$\frac{1}{f}=\frac{1}{10}-\frac{1}{-10} or f=+5$
Further, $\Delta u=0.1$
$\Delta v=0.1$(from the graph)
Now, differentiating the lens formula, we have
$\frac{\Delta f}{f^2}=\frac{\Delta v}{v^2}+\frac{\Delta u}{u^2}$
$\Delta f=(\frac{\Delta v}{v^2}+\frac{\Delta u}{u^2})$
Substituting the values, we have
$\Delta f=(\frac{0.1}{10^2}+\frac{0.1}{10^2})(5)^2 =0.05$
$\therefore f\pm \Delta f=5\pm 0.05$
