A combination of two thin lenses with focal lengths $f_{1}$ and $f_{2}$ respectively forms an image of distant object at distance 60 cm when lenses are in contact. The position of this image shifts by 30 cm towards the combination when two lenses are separated by 10 cm. The corresponding values of $f_{1}$ and $f_{2}$ are

30 cm, - 60 cm

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20 cm, - 30 cm

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15 cm, - 20 cm

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12 cm, – 15 cm

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Solution

The correct option is B 20 cm, - 30 cm
Initially F =60 cm (Focal length of combination)
Hence by using $\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} \Rightarrow \frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{1}{60} \Rightarrow \frac{f_{1} f_{2}}{f_{1} + f_{2}} . . . . (i)$
Finally by using $\frac{1}{F^{'}} = \frac{1}{f_{1}} + \frac{1}{f_{2}} - \frac{d}{f_{1} f_{2}}$ where F'=30 cm and d =10 cm $\Rightarrow \frac{1}{30} = \frac{1}{f_{1}} + \frac{1}{f_{2}} - \frac{10}{f_{1} f_{2}} . . . . (i i)$
From equations (i) and (ii) $f_{1} . f_{2} = - 600$ .
From equation (i) $f_{1} + f_{2} = - 10 . . . . . . . . . . . (i i i)$ .
Also,difference of focal lengths can written as $f_{1} - f_{2} = \sqrt{(f_{1} + f_{2})^{2} - 4 f_{1} f_{2}} \Rightarrow f_{1} - f_{2} = 50 . (i v)$
From (iii) ´ (iv) $f_{1} = 20$ and $f_{2} = - 30$

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