A microscope consists of two convex lenses of focal lengths 2cm and 6.25cm placed 15cm apart. Where must the object be placed so that final virtual image is at a distance of 25cm from eye

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2cm

1.5cm

3cm

2.5cm

answer is A.

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Detailed Solution

For eyepiece,

$\frac{1}{v_{e}} - \frac{1}{u_{e}} = \frac{1}{f_{e}}$

$\Rightarrow |u_{e}| = \frac{f_{e} D}{f_{e} + D}$

Also, $|v_{0}| + |u_{e}| = L$

$\Rightarrow v_{0} + \frac{f_{e} D}{f_{e} + D} = L$

$v_{0} = L - \frac{f_{e} D}{f_{e} + D} = 15 - \frac{6.25 \times 25}{31.25} \Rightarrow v_{0} = 10$

For objective lens,

$\frac{1}{v_{0}} - \frac{1}{u_{0}} = \frac{1}{f_{0}}$

$\frac{1}{10} - \frac{1}{u_{0}} = \frac{1}{2}$

$\Rightarrow u_{0} = - \frac{10}{4} = - 2.5 cm$

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