Question

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 -

• A. $\left(\text{0.50 }\pm ;\text{ 0.05}\right)\text{cm}$
• B. $\left(\text{5.00 }\pm ;\text{ 0.05}\right)\text{cm}$
• C. $\left(\text{0.50 }\pm ;\text{ 0.10}\right)\text{cm}$
• D. $\left(\text{5.00 }\pm ;\text{ 0.10}\right)\text{cm}$

Answer 1

Answer: (B)

$2f=10$
$f=5cm$
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\frac{df}{f^2}=\frac{dv}{v^2}+\frac{du}{u^2}$
$df=f^2\left(\frac{dv}{v^2}+\frac{du}{u^2}\right)$
$u=10,v=10,du=\frac{1}{10},dv=\frac{1}{10}$
$\text{df}={\left(\text{f}\right)}^2\left(\frac{\text{dv}}{{\left(\text{v}\right)}^2}+\frac{\text{du}}{{\left(\text{u}\right)}^2}\right)$
$u=10,v=10,\text{du}=\frac{1}{10}$ , $\text{dv}=\frac{1}{10}$
(as there are $10$ division in $1cm$ scale, least count is $0.1$ i.e., considered as error)
So $\text{df}=25\left(\frac{1}{10{\left(10\right)}^2}+\frac{1}{10{\left(10\right)}^2}\right)=25\left[\frac{2}{10\times 100}\right]$
$df=0.05s$ of $=(5.00\pm ;0.05)cm$.