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Magnification in Thin Lens

When the obje...

Question

When the object is placed at $4 cm$ from the objective lens of the compound microscope, the final image is formed at the position of the object which is the least distance of distinct vision $(24 cm)$ . If the focal length of the eyepiece is $8 cm$ then the focal length of the objective lens is -

3 cm

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2 cm

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4 cm

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5 cm

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Solution

The correct option is A $3 cm$

Given:
$u_{o} = - 4 cm$
$v_{e} = - 24 cm$
$f_{e} = 8 cm$

Applying lens formula for eyepiece,

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

$\Rightarrow \frac{1}{8} = \frac{1}{(- 24)} - \frac{1}{u_{e}}$

$\Rightarrow u_{e} = - 6 cm$

Since, the final image is formed at the position of the object, the length of the tube is given by,

$L = | v_{e} | - | u_{o} |$

$\Rightarrow L = 24 - 4 = 20 cm$

The length of the tube is also given by,

$L = | v_{o} | + | u_{e} |$

$∴ | v_{o} | = 20 - 6 = 14 cm$

Further, applying lens formula for objective lens,

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

$\Rightarrow \frac{1}{f_{0}} = \frac{1}{14} - \frac{1}{(- 4)}$

$\Rightarrow f_{o} \approx 3 cm$

Hence, option $(A)$ is the correct answer.

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