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)

1.0 m

Answer 2

Let a large convex lens is placed between two walls at a distance x from wall on which an electric bulb is fixed. Using lens formula,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
Putting $u=x$ and $v=4-x$
$\therefore$ $\frac{1}{f}=\frac{1}{4-x}-\frac{1}{-x}$
Or $\frac{1}{f}=\frac{x+4-x}{(4-x)(x)}$
Or $\frac{1}{f}=\frac{4}{(4-x)(x)}$
Or $f=\frac{(4-x)(x)}{4}$ .. (i)
Now magnification, $m=\frac{v}{u}=\frac{4-x}{x}$
or $1=\frac{4-x}{x}$
or $x=4-x$ or
$2x=4$ or $x=2m$
Hence, from equation (i),
$f=\frac{(4-2)(2)}{4}=\frac{2\times 2}{4}=1m$