Question

The graph between object distance $u$ and image distance $v$ from a convex lens is given. The correct option option, from the following, is

• A. $f= 5\pm0\cdot 1$
• B. $f \, = 5\pm0\cdot 05$
• C. $f= 0\cdot 5\pm0\cdot 1$
• D. $f= 0\cdot 5\pm0\cdot 05$

Answer 1

Answer: (B)

From the lens formula :
We have $\frac{1}{\textit{f}}=\frac{1}{\upsilon}-\frac{1}{\textit{u}}$
$\frac{1}{\textit{f}}=\frac{1}{1 0}-\frac{1}{- 1 0}$
or, $\textit{f}=+5$
Further, $\Delta \textit{u}=0\cdot 1$
and $\Delta \upsilon=0\cdot 1$ ( from the graph )
Now, differentiating the lens formula, we have
$\frac{\Delta \textit{f}}{\textit{f}^{2}}=\frac{\Delta \upsilon}{\upsilon^{2}}+\frac{\Delta \textit{u}}{\textit{u}^{2}}$
or, $\Delta f=\left(\frac{\Delta v}{v^2}+\frac{\Delta u}{u^2}\right) \quad f^2$
Substituting the values, we get
$\Delta \textit{f}=\left(\frac{0 \cdot 1}{1 0^{2}} + \frac{0 \cdot 1}{1 0^{2}}\right)\left(5\right)^{2}=0\cdot 05$

$∴ \, \, \textit{f}\pm\Delta \textit{f}=5\pm0\cdot 05$