The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3𝑚 away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?

Step 1: Draw the diagram.

Step 2: Find the value of image distance.
Formula Used: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
Suppose the object distance is u,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
For convex lens, u is negative so by lens formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
As seen from the diagram
$u+v=d$
$v=d-u$
Put the value of v in the lens formula,
$\frac{1}{d-u}+\frac{1}{u}=\frac{1}{f}$
$\therefore \frac{u+d-u}{u(d-u)}=\frac{1}{f}$
$f=\frac{u(d-u)}{d}.......\left(1\right)$
For maximum focal length,
$\frac{df}{du}=0$
$\frac{df}{du}=\frac{d-2u}{d}=0$
$u=\frac{d}{2}$
Put the value of $u$ in (1) for maximum focal length,
$f_{max}=\frac{d}{4}$
$=\frac{3}{4}=0.75\text{cm}$
Hence, for the required purpose, the maximum possible focal length of the convex lens is $0.75m$.
Final Answer: $f_{max}=0.75m$.