The focal length of the objective and eye len | vedclass

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(C) For a microscope,the magnifying power $M$ when the final image is at infinity is given by $M = m_o 	imes m_e$. Here,$m_o$ is the magnification of

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$24$

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(C) For a microscope,the magnifying power $M$ when the final image is at infinity is given by $M = m_o 	imes m_e$. Here,$m_o$ is the magnification of the objective lens and $m_e$ is the magnification of the eyepiece. The magnification of the objective is $m_o = \frac{v_o}{u_o}$. Using the lens formula for the objective: $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$. Given $f_o = 4 \ cm$ and $u_o = -4.5 \ cm$,we have $\frac{1}{v_o} = \frac{1}{4} - \frac{1}{4.5} = \frac{1}{4} - \frac{2}{9} = \frac{9-8}{36} = \frac{1}{36}$. So,$v_o = 36 \ cm$. The magnification of the objective is $m_o = \frac{v_o}{u_o} = \frac{36}{-4.5} = -8$. The magnification of the eyepiece for the final image at infinity is $m_e = \frac{D}{f_e} = \frac{24}{8} = 3$. The total magnifying power is $M = |m_o 	imes m_e| = |-8 	imes 3| = 24$.

The focal length of the objective and eye lens of a microscope are $4 \ cm$ and $8 \ cm$ respectively. If the least distance of distinct vision is $24 \ cm$ and the object distance is $4.5 \ cm$ from the objective lens,then the magnifying power of the microscope will be: (Final image is at infinity)

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