The height of a candle flame is 5 cm. A lens produces an image of this flame 15cm high on a screen. The candle is then moved over a distance 1.5 cm away from the lens and a sharp of the flame 10cm high is obtained after shifting the screen. Determine the focal length of the lens (in cm)

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answer is 9.

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Detailed Solution

For convex lens forming real image
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
$m = \frac{15}{5} = \frac{v}{u}$ or $3 u = v$
$∴ \frac{1}{3 u} + \frac{1}{u} = \frac{1}{f}$ or $u = \frac{4 f}{3}$ and $v = 4 f$
In the second position $u^{'} = u + 15 = \frac{4 f}{3} + 1.5$
$m^{'} = \frac{v^{'}}{u^{'}} = \frac{10}{5}$
$v^{'} = 2 u^{'} = 2 (\frac{4 f}{3} + 1.5) = \frac{8 f}{3} + 3$
$\frac{1}{v^{1}} + \frac{1}{u^{'}} = \frac{1}{f^{'}}$
Or $\frac{1}{(\frac{8 f}{3}) + 3} + \frac{1}{(\frac{4 f}{3}) + 1.5} = \frac{1}{f}$
$∴ f = 9 c m o r - 1.125 c m$
Only $f = 9 c m$ is correct answer. The other value is absurd

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