Question 423: Ray Optics

Question

Graph of position of image versus position of point object from a convex lens is shown. Then, the focal length of the lens is
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Accepted Answer

Given : $v-u$ graph of a convex lens.
from
$ \begin{aligned} & \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \ & \frac{1}{f}=\frac{1}{10}-\frac{1}{-10}=\frac{1}{10}+\frac{1}{10} \end{aligned} $
$ \begin{aligned} &\begin{gathered} \frac{1}{f}=\frac{2}{10}=\frac{1}{5} \ f=5 \mathrm{~cm} \end{gathered}\ &	ext { On differentiation of the given formula of } f 	ext {. }\ &\begin{aligned} -\frac{d f}{f^2} & =\frac{-d v}{v^2}-\frac{d u}{u^2} \ \frac{\Delta f}{f^2} & =\frac{\Delta v}{v^2}+\frac{\Delta u}{u^2} \ \frac{\Delta f}{5^2} & =\frac{0.1}{10^2}+\frac{0.1}{10^2}=\frac{0.2}{100} \ \Delta f & =\frac{0.2 	imes 25}{100}=0.05 \mathrm{~cm} \ f & =5 \pm 0.05 \mathrm{~cm} \end{aligned} \end{aligned} $

Answer

$0.50 \pm 0.05 \mathrm{~cm}$

Answer

$0.50 \pm 0.10 \mathrm{~cm}$

Answer

$5.00 \pm 0.05 \mathrm{~cm}$

Answer

$5.00 \pm 0.10 \mathrm{~cm}$

Ray Optics

Graph of position of image versus position of point object from a convex lens is shown. Then, the focal length of the lens is

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