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Q.
Graph of position of image vs position of a point object from a convex lens is shown in the figure. The focal length of the lens is -
NTA AbhyasNTA Abhyas 2020
Solution:
$2 f =10$
$f =5 cm$
$\frac{1}{ f }=\frac{1}{ v }-\frac{1}{ u }$
$\frac{ df }{ f ^{2}}=\frac{ dv }{ v ^{2}}+\frac{ du }{ u ^{2}}$
$df = f ^{2}\left(\frac{ dv }{ v ^{2}}+\frac{ du }{ u ^{2}}\right)$
$u =10, v =10, du =\frac{1}{10}, dv =\frac{1}{10}$
$\text{df} = \left(\text{f}\right)^{2} \left(\frac{\text{dv}}{\left(\text{v}\right)^{2}} + \frac{\text{du}}{\left(\text{u}\right)^{2}}\right)$
$u = 10, v = 10, \text{du} = \frac{1}{1 0}$ , $\text{dv} = \frac{1}{1 0}$
(as there are $10$ division in $1\, cm$ scale, least count is $0.1$ i.e., considered as error)
So $\text{df} = 2 5 \left(\frac{1}{1 0 \left(1 0\right)^{2}} + \frac{1}{1 0 \left(1 0\right)^{2}}\right) = 2 5 \left[\frac{2}{1 0 \times 1 0 0}\right]$
$df = 0.05\, s$ of $=(5.00 \pm; 0.05) cm$.
