A compound microscope has an eyepiece of foca | vedclass

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(A) For the objective lens,the lens formula is $\frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o}$.Given $f_o = 4 \, cm$ and $u_o = -5 \, cm$,we have $\fr

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$12$

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(A) For the objective lens,the lens formula is $\frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o}$.Given $f_o = 4 \, cm$ and $u_o = -5 \, cm$,we have $\frac{1}{v_o} = \frac{1}{4} + \frac{1}{-5} = \frac{5-4}{20} = \frac{1}{20}$.Thus,$v_o = 20 \, cm$.The magnification of a compound microscope when the final image is at the least distance of distinct vision $(D)$ is given by $M = \frac{v_o}{|u_o|} \left( 1 + \frac{D}{f_e} ight)$.Substituting the values: $M = \frac{20}{5} \left( 1 + \frac{20}{10} ight)$.$M = 4 	imes (1 + 2) = 4 	imes 3 = 12$.

$A$ compound microscope has an eyepiece of focal length $10 \, cm$ and an objective of focal length $4 \, cm$. Calculate the magnification,if an object is kept at a distance of $5 \, cm$ from the objective so that the final image is formed at the least distance of distinct vision $(20 \, cm)$.

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$A$ compound microscope has a magnifying power of $30$. The focal length of its eyepiece is $5 \ cm$. Assuming the final image is at the least distance of distinct vision $(25 \ cm)$,the magnification produced by the objective is:

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