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(D) Given: Focal length of objective $f_o = 1 \ cm$,focal length of eye-piece $f_e = 6 \ cm$,tube length $L = 30 \ cm$,and least distance of distinct

Vedclass · Accepted Answer

$125$

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(D) Given: Focal length of objective $f_o = 1 \ cm$,focal length of eye-piece $f_e = 6 \ cm$,tube length $L = 30 \ cm$,and least distance of distinct vision $D = 25 \ cm$.For the eye-piece,the image is formed at $D = 25 \ cm$. Using the lens formula $\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e}$,where $v_e = -D = -25 \ cm$:$\frac{1}{-25} - \frac{1}{u_e} = \frac{1}{6} \implies \frac{1}{u_e} = -\frac{1}{25} - \frac{1}{6} = -\frac{31}{150} \implies u_e = -\frac{150}{31} \approx -4.84 \ cm$.The magnitude is $|u_e| = 4.84 \ cm$.The magnification $M$ for a compound microscope is given by $M = m_o 	imes m_e = \left(\frac{L - f_o - |u_e|}{f_o}ight) \left(1 + \frac{D}{f_e}ight)$.Substituting the values: $M = \left(\frac{30 - 1 - 4.84}{1}ight) \left(1 + \frac{25}{6}ight) = (24.16) 	imes (1 + 4.166) = 24.16 	imes 5.166 \approx 124.8$.Rounding to the nearest integer,the magnification is $125$.

$A$ microscope has an objective of focal length $1 \ cm$ and an eye-piece of focal length $6 \ cm$. If the tube length is $30 \ cm$ and the image is formed at the least distance of distinct vision,what is the magnification produced by the microscope? Take $D = 25 \ cm$.

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