A compound microscope has a magnifying power | vedclass

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(A) The magnifying power of a compound microscope is given by $M = m_o 	imes m_e$.The magnification of the eyepiece $(m_e)$ when the final image is a

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$5$

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(A) The magnifying power of a compound microscope is given by $M = m_o 	imes m_e$.The magnification of the eyepiece $(m_e)$ when the final image is at the least distance of distinct vision $(D = 25 \, cm)$ is given by the formula $m_e = 1 + \frac{D}{f_e}$.Given $f_e = 5 \, cm$ and $D = 25 \, cm$,we calculate $m_e$:$m_e = 1 + \frac{25}{5} = 1 + 5 = 6$.Given the total magnifying power $M = 30$,we substitute the values into the formula $M = m_o 	imes m_e$:$30 = m_o 	imes 6$.Solving for $m_o$:$m_o = \frac{30}{6} = 5$.Thus,the magnification produced by the objective is $5$.

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

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