The focal lengths of the objective and the ey | vedclass

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(B) Given: Focal length of objective $f_o = 1 \, cm$,focal length of eyepiece $f_e = 5 \, cm$,and object distance $u_o = -1.1 \, cm$.First,find the im

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$50$

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(B) Given: Focal length of objective $f_o = 1 \, cm$,focal length of eyepiece $f_e = 5 \, cm$,and object distance $u_o = -1.1 \, cm$.First,find the image distance $v_o$ for the objective lens using the lens formula: $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$.$\frac{1}{v_o} = \frac{1}{f_o} + \frac{1}{u_o} = \frac{1}{1} - \frac{1}{1.1} = \frac{1.1 - 1}{1.1} = \frac{0.1}{1.1} = \frac{1}{11}$.So,$v_o = 11 \, cm$.The magnification of the objective is $m_o = \frac{v_o}{u_o} = \frac{11}{-1.1} = -10$.Since the final image is formed at infinity,the eyepiece acts as a simple magnifier with magnification $m_e = \frac{D}{f_e}$,where $D = 25 \, cm$ (least distance of distinct vision).$m_e = \frac{25}{5} = 5$.The total magnifying power $M = m_o 	imes m_e = (-10) 	imes 5 = -50$.Taking the magnitude,the magnifying power is $50$.

The focal lengths of the objective and the eyepiece of a compound microscope are $1 \, cm$ and $5 \, cm$ respectively. An object is placed at a distance of $1.1 \, cm$ from the objective. If the final image is formed at infinity,the magnifying power of the microscope is:

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