The length of the compound microscope is 15 | vedclass

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(B) For a compound microscope,the magnifying power $M$ for a relaxed eye is given by $M = m_o 	imes m_e$,where $m_o = \frac{v_o}{u_o}$ and $m_e = \fr

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$1.5$

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(B) For a compound microscope,the magnifying power $M$ for a relaxed eye is given by $M = m_o 	imes m_e$,where $m_o = \frac{v_o}{u_o}$ and $m_e = \frac{D}{f_e}$.Given $M = 25$,$f_e = 6 \ cm$,and $D = 25 \ cm$ (standard near point).$m_e = \frac{25}{6} \approx 4.167$.So,$m_o = \frac{M}{m_e} = \frac{25}{25/6} = 6$.Since $m_o = \frac{v_o}{u_o} = 6$,we have $v_o = 6u_o$.The length of the microscope tube $L = v_o + f_e = 15 \ cm$.Substituting $v_o = 6u_o$,we get $6u_o + 6 = 15$.$6u_o = 9 \implies u_o = 1.5 \ cm$.

The length of the compound microscope is $15 \ cm$. The magnifying power for a relaxed eye is $25$. If the focal length of the eye lens is $6 \ cm$,then the object distance for the objective lens will be: (in $cm$)

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