Question

a compound microscope has an eye piece of focal length 10cm and an objective of focal length 4cm the magnification of an object is placed at a distance of 5cm from the objective so that the final imageis formed at the least distance vision (d=20) is:​

Answer 1

Answer:

12 is the correct answer bro

Answer 2

☞︎︎︎ Solution :

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• Fo = 4cm

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• Fe = 10cm

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• Uo = 5cm

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$\large \rightarrow \rm \: m = \dfrac{vo}{uo} \left(1 + \dfrac{d}{fe} \right)$

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$\large \: \: \: \: : \implies \: \rm\dfrac{1}{f} = \dfrac{1}{vo} - \dfrac{1}{uo}$

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$\large \: \: \: \: : \implies \: \rm\dfrac{1}{vo} = \dfrac{1}{f} + \dfrac{1}{uo}$

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$\large \: \: \: \: : \implies \: \rm\dfrac{1}{vo} = \dfrac{1}{4} + \dfrac{1}{5}$

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$\large \: \: \: \: : \implies \: \rm\dfrac{1}{vo} = \dfrac{1}{20}$

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$\large \rm \therefore \boxed{ \underline{\rm vo = 20cm}}$

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$\large \rightarrow \rm \: m = \dfrac{vo}{uo} \left(1 + \dfrac{d}{fe} \right)$

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$\large \: \: \: \: : \implies \rm \: m = \dfrac{20}{5} \left(1 + \dfrac{20}{10} \right)$

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$\large \: \: \: \: : \implies \rm \: m = \dfrac{ \cancel{20} \: 4}{ \cancel5} \left(1 + \dfrac{2}{1} \right)$

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$\large \: \: \: \: : \implies \rm \: m = 4(1 + 2)$

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$\: \: \large \rm \therefore \: \: \boxed{ \underline{\rm m= 12cm}}$