Two thin lenses, when in contact, produce a combination of power +10 dioptre. When they are 0.25 m apart, the power is reduced to +6 diptre. The respective power of the lenses in diopter, are

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

1 and 9

5 each

2 and 8

4 and 6

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

When in contact, we have

$\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} \Rightarrow 10 = \frac{1}{f_{1}} + \frac{1}{f_{2}} {∵ P = \frac{1}{F} = 10}$

$\Rightarrow f_{1} + f_{2} = 10 f_{1} f_{2} ............. (1)$

When at a separation of 0.25 m, we have
$\frac{1}{F^{1}} = \frac{1}{f_{1}} + \frac{1}{f_{2}} - \frac{x}{f_{1} f_{2}} \Rightarrow 6 = \frac{1}{f_{1}} + \frac{1}{f_{2}} - \frac{0.25}{f_{1} f_{2}}$
$\Rightarrow 6 = 10 - \frac{0.25}{f_{1} f_{2}} \Rightarrow f_{1} f_{2} = \frac{1}{16} ............... (2)$
From (1), we get $f_{1} + f_{2} = \frac{10}{16} = \frac{5}{8}$
Now, from (2) we get $f_{1} (\frac{5}{8} - f_{1}) = \frac{1}{16}$
$\Rightarrow 16 f_{1}^{2} - 10 f_{1} + 1 = 0 \Rightarrow 16 f_{1}^{2} - 8 f_{1} - 2 f_{1} + 1 = 0$