Explanation

(b) Let a large convex lens is placed between two walls at a distance $x$ from wall on which an electric bulb is fixed.
Using
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$= \frac{1}{4 - x} - \frac{1}{- x}\ \ \ \ \ \ \ \ \ (\because u = - x,\ v = 4 - x)$
$= \frac{x + 4 - x}{(4 - x)(x)}$
$= \frac{4}{(4 - x)(x)}$
$or\ f = \frac{(4 - x)x}{4}\ \ \ \ \ \ \ \ldots\ldots(i)$
Now magnification $m = \frac{v}{u} = \frac{4 - x}{x}$
$\Longrightarrow \ 1 = \frac{4 - x}{x}$
$\Longrightarrow x = 4 - x$
$\Longrightarrow 2x = 4$
$\Longrightarrow x = 2m$
From eq. (i)
$f = \frac{(4 - 2)(2)}{4} = \frac{2 \times 2}{4} = 1m$