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(B) Given:Focal length of objective,$f_o = 4\,cm$Focal length of eyepiece,$f_e = 8\,cm$Tube length,$L = 30\,cm$For normal adjustment,the image formed

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$14.06$

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(B) Given:Focal length of objective,$f_o = 4\,cm$Focal length of eyepiece,$f_e = 8\,cm$Tube length,$L = 30\,cm$For normal adjustment,the image formed by the objective lies at the focal point of the eyepiece. Thus,the distance of the image from the objective is $v_o = L - f_e = 30 - 8 = 22\,cm$.Using the lens formula for the objective: $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$$\frac{1}{22} - \frac{1}{u_o} = \frac{1}{4} \implies \frac{1}{u_o} = \frac{1}{22} - \frac{1}{4} = \frac{2 - 11}{44} = -\frac{9}{44}$$u_o = -\frac{44}{9}\,cm$The magnification of the objective is $m_o = \frac{v_o}{u_o} = \frac{22}{-44/9} = -\frac{22 	imes 9}{44} = -4.5$The magnification of the eyepiece for normal adjustment is $m_e = \frac{D}{f_e} = \frac{25}{8} = 3.125$The total magnifying power is $M = m_o 	imes m_e = -4.5 	imes 3.125 = -14.0625$.Taking the magnitude,$M \approx 14.06$.

In a compound microscope,the focal length of the objective and eyepiece lens is $4\,cm$ and $8\,cm$ respectively. If the length of the tube is $30\,cm$,then the magnifying power for normal adjustment is:

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