Question

The height of a candle flame is 5 cm. A lens produces an image of this flame 15 m height on a screen. Without touching the
lens, the candle is moved over a distance of 1 = 1.5 cm away from the lens and a sharp image of the flame 10 cm high is
obtained again after shifting the screen. The focal length of the lens is​

Answer 1

Explanation:

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Answer 2

Given,

The height of candle = 5cm

The lens produced height of flame image on the screen = 15m

Candle moved over a distance (l) = 1.5cm

The height of the second image of flame after moving away = 10cm

To Find: The focal length of the lens?

Explanation:

Applying the formula of a lens to both cases, we obtain,

                             $\frac{1}{a_{1} } +\frac{1}{b_{1} } =\frac{1}{f}$

                   and,

                             $\frac{1}{a_{2} } +\frac{1}{b_{2} } =\frac{1}{f}$

According to the initial condition,

                              $a_{2}=a_{1} +l$

                  and,

                              $\frac{b_{1} }{a_{1} } = k_{1} = 3$

Magnification in the first case,

                              $\frac{b_{2} }{a_{2} } = k_{2} = 2$

Magnification in the second case,

                              $f = \frac{k_{1} k_{2} }{k_{1} -k_{2} } l$

                              $f = \frac{6}{1}$×$1.5$

                              $f = 9 cm$

Hence the focal length of the lens is 9cm.