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(A) Given: Focal length of objective $f_o = 0.95 \ cm$,focal length of eyepiece $f_e = 5 \ cm$,distance between lenses $L = 20 \ cm$,and final image d

Vedclass · Accepted Answer

$94$

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(A) Given: Focal length of objective $f_o = 0.95 \ cm$,focal length of eyepiece $f_e = 5 \ cm$,distance between lenses $L = 20 \ cm$,and final image distance $v_e = -25 \ cm$.First,find the object distance for the eyepiece $(u_e)$ using the lens formula: $\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e}$.$\frac{1}{-25} - \frac{1}{u_e} = \frac{1}{5} \Rightarrow \frac{1}{u_e} = -\frac{1}{25} - \frac{1}{5} = -\frac{6}{25} \Rightarrow u_e = -\frac{25}{6} \approx -4.17 \ cm$.The distance between lenses is $L = v_o + |u_e| = 20 \ cm$.$v_o = 20 - 4.17 = 15.83 \ cm$.Magnification $m = m_o 	imes m_e = (\frac{v_o}{u_o}) 	imes (1 + \frac{D}{f_e})$.Using $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$,we get $\frac{1}{u_o} = \frac{1}{v_o} - \frac{1}{f_o} = \frac{1}{15.83} - \frac{1}{0.95} \approx 0.063 - 1.053 = -0.99$.$m_o = \frac{v_o}{u_o} = 15.83 	imes (-0.99) \approx -15.67$.$m_e = (1 + \frac{25}{5}) = 6$.Total magnification $m = -15.67 	imes 6 \approx -94$. The magnitude is $94$.

In a compound microscope,the objective lens and eyepiece have focal lengths of $0.95 \ cm$ and $5 \ cm$ respectively,and are kept at a distance of $20 \ cm$. The final image is formed at a distance of $25 \ cm$ from the eyepiece. Calculate the magnifying power.

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