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 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 $