Question

The graph shows the relationship between object distance u and image distance v for an equiconvex lens. The focal length of the convex lens is

• A. $0.50\pm 0.05cm$
• B. $0.050\pm 0.10cm$
• C. $5.00+0.05cm$
• D. $5.00+0.10cm$

Answer 1

Answer: (C)

$5.00+0.05cm$

$\begin{array}{c}\text{From graph, when}u=-10cm\\ v=10cm\text{. when an object is}\\ \text{placed at the center of}\\ \text{curvature (c) of a convex}\\ \text{lens, its image is formed}\\ \text{at center of curvature.}\\ \text{i.e.}R=2f=|u|=|v|\\ \Rightarrow f=\frac{v}{2}=\frac{10cm}{2}=5cm\end{array}$

$\begin{array}{c}\text{According to lens equation}\\ \frac{1}{f}=\frac{1}{u}+\frac{1}{v}-1\\ \text{Maximum error in}f\text{is}\\ \text{given}by\text{differentiating}\\ \text{(1).}\\ \frac{\Delta f}{f^2}=\frac{\Delta v}{v^2}+\frac{\Delta u}{u^2}\end{array}$

$\text{or}\begin{array}{cc}\frac{\Delta f}{f^2}=\frac{0.1}{(10{)}^2}+\frac{0.1}{{10}^2}& \\ =& \frac{0.2}{100}[\begin{array}{c}\Delta u=\Delta v=0.1\\ \text{Least count of}\\ \text{the scale}=0.1\end{array}\\ \Rightarrow \begin{array}{cc}\Delta f& =f^2\times \frac{0.2}{100}\\ & =5^2\times \frac{0.2}{100}=0.05cm\end{array}& \end{array}$

$\begin{array}{c}\text{Focal length of the lens}\\ =f\pm \Delta f=5.00\pm 0.05\end{array}$

$\begin{array}{c}\text{Taking maximum error,}\\ f=5.00+0.05cm\end{array}$