Question

The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall $4\,m$ away by means of a large convex lens. The maximum possible focal length of the lens required for this purpose will be

• A. 0.5 m
• B. 1.0 m
• C. 1.5 m
• D. 2.0 m

Answer 1

Answer: (B)

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