Search any question & find its solution
Question:
Answered & Verified by Expert
The graph between object distance $u$ and image distance $v$ for a lens is given below. The focal length of the lens is
Options:
Solution:
2338 Upvotes
Verified Answer
The correct answer is:
$5 \pm 0.05$
$5 \pm 0.05$
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, $\Delta u=0.1$ and $\Delta v=0.1$
(from the graph)
Now, differentiating the lens formula we have,
$$
\frac{\Delta f}{f^2}=\frac{\Delta v}{v^2}+\frac{\Delta u}{u^2} \text { or } \Delta f=\left(\frac{\Delta v}{v^2}+\frac{\Delta u}{u^2}\right) f^2
$$
Substituting the values we have,
$$
\Delta f=\left(\frac{0.1}{10^2}+\frac{0.1}{10^2}\right)(5)^2 \Rightarrow f \pm \Delta f=5 \pm 0.05
$$
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ we have, $\frac{1}{f}=\frac{1}{10}-\frac{1}{-10}$ or $f=+5$
Further, $\Delta u=0.1$ and $\Delta v=0.1$
(from the graph)
Now, differentiating the lens formula we have,
$$
\frac{\Delta f}{f^2}=\frac{\Delta v}{v^2}+\frac{\Delta u}{u^2} \text { or } \Delta f=\left(\frac{\Delta v}{v^2}+\frac{\Delta u}{u^2}\right) f^2
$$
Substituting the values we have,
$$
\Delta f=\left(\frac{0.1}{10^2}+\frac{0.1}{10^2}\right)(5)^2 \Rightarrow f \pm \Delta f=5 \pm 0.05
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.