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When the obje...

Question

When the object is placed 4 cm from the objective of microscope, the final image formed coincides with the object. The final image is at the least distance of distinct vision (24 cm). If magnifying power of microscope is 15, the focal lengths of objective and eye-piece is

f_{o} =

3.125 cm,

f_{e} =

7.5 cm

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f_{o} =

2 cm,

f_{e} =

3 cm

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f_{o} =

3 cm ,

f_{e} =

4 cm

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f_{o} =

4 cm ,

f_{e} =

5 cm

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Solution

The correct option is A $f_{o} =$ 3.125 cm, $f_{e} =$ 7.5 cm
$u_{0} = 4 c m$

$v_{e} = 24 c m$

$L = 20 c m$

$M = 15$

$f_{0} = ?$
$f_{e} = ?$

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

$M_{D} = \frac{v_{0}}{4} (1 + \frac{25}{f_{e}})$

$\frac{60 f_{e}}{(f_{e} + 25)} = v_{0}$

$v_{0} = \frac{60 \times 7.5}{31.5} = 14.285$

$\frac{7}{100} - \frac{1}{4} = \frac{1}{f_{0}}$

$f_{0} = 3.125 c m$

$20 = \frac{60 f_{e}}{(f_{e} + 24)} + \frac{24 f_{e}}{(f_{e} + 24)}$

$20 = \frac{84 f_{e}}{f_{e} + 24}$

$20 f_{e} + 480 = 84 f_{e}$

$480 = 64 f_{e}$

$f_{e} = \frac{480}{64}$ $= 7.5 c m$

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