A bulb is located on a wall. Its image is to be obtained on a parallel walls with the help of a convex lens.…
- only $\frac{d}{4}$
- only $\frac{d}{2}$
- more than $\frac{d}{4}$ but less than $\frac{d}{2}$
- less than or equal to $\frac{d}{4}$
Solution
Since $v+u=d \Rightarrow v=d-u$
From lens formula
$\begin{aligned}
\frac{1}{f} & =\frac{1}{v}-\frac{1}{u} \\
\Rightarrow \quad \frac{1}{f} & =\frac{1}{d-u}-\frac{1}{-u} \\
& =\frac{d}{(d-u) u} \\
\Rightarrow \quad f & =\frac{d u-u^2}{d} \ldots(\mathrm{i})
\end{aligned}$
Differentiating eq. (i) we have
$\frac{d f}{d u}=\frac{1}{d}(d-2 u)$
for maximum value of $f_1=\frac{d f}{d u}=0$
$\begin{aligned}
& \Rightarrow \frac{1}{d}(d-2 u)=0 \Rightarrow d=24 \\
& \Rightarrow u=\frac{d}{2}
\end{aligned}$
putting these value in eq. (i)
$f=\frac{d}{4}$Asked in: NEET 2002