Two thin lenses, when in contact, produce a combination of power $+ 10 D$ . When they are $0.25 m$ apart, the power reduces to $+ 6 D$ . The focal lengths of the lenses (in m) are

0.124

and

0.5

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0.125

and

0.125

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0.15

and

0.75

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0.125

and

0.75

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Solution

The correct option is A $0.124$ and $0.5$
When lenses are in contact
$P = \frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} \Rightarrow 10 = \frac{1}{f_{1}} + \frac{1}{f_{2}}$
When They are distance $d$ apart
$P^{'} = \frac{1}{F^{'}} = \frac{1}{f_{1}} + \frac{1}{f_{2}} - \frac{d}{f_{1} f_{2}} \Rightarrow 6 = \frac{1}{f_{1}} + \frac{1}{f_{2}} - \frac{0.25}{f_{1} f_{2}}$
From Equation (i) and (ii) $f_{1} f_{2} = \frac{1}{16}$
From equation (i) and (iii) $f_{1} + f_{2} = \frac{5}{8}$
Also $(f_{1} - f_{2})^{2} = (f_{1} + f_{2})^{2} - 4 f_{1} f_{2}$
Hence $(f_{1} - f_{2})^{2} = {(\frac{5}{8})}^{2} - 4 \times \frac{1}{16} = \frac{9}{64}$
$\Rightarrow f_{1} - f_{2} = \frac{3}{8}$
On solving $(i v)$ and $(v) f_{1} = 0.5 m$ and $f_{2} = 0.125 m$

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