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(A) For a compound microscope,the total magnification $M$ is given by $M = m_o 	imes m_e$. Since the final image is formed at the least distance of d

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$4$

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(A) For a compound microscope,the total magnification $M$ is given by $M = m_o 	imes m_e$. Since the final image is formed at the least distance of distinct vision $(D = 25 \ cm)$,the magnification of the eyepiece is given by $m_e = (1 + D/f_e)$. Given $M = 24$,$f_e = 5 \ cm$,and $D = 25 \ cm$: $m_e = 1 + \frac{25}{5} = 1 + 5 = 6$. Now,substituting the values into the total magnification formula: $24 = m_o 	imes 6$. Therefore,$m_o = \frac{24}{6} = 4$.

$A$ compound microscope produces a magnification of $24$. The focal length of the eyepiece is $5 \ cm$. The final image is formed at the least distance of distinct vision. The magnification produced by the objective is

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