A microscope consists of an objective of foca | vedclass

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(B) For the eyepiece,the image distance $v_e = -25 \ cm$ and focal length $f_e = 5 \ cm$. Using the lens formula $\frac{1}{v_e} - \frac{1}{u_e} = \fra

Vedclass · Accepted Answer

$2.7$

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(B) For the eyepiece,the image distance $v_e = -25 \ cm$ and focal length $f_e = 5 \ cm$. Using the lens formula $\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e}$:$\frac{1}{-25} - \frac{1}{u_e} = \frac{1}{5} \Rightarrow \frac{1}{u_e} = -\frac{1}{25} - \frac{1}{5} = -\frac{6}{25} \Rightarrow u_e = -\frac{25}{6} \ cm$.The distance of the image formed by the objective from the objective is $v_0 = L - |u_e| = 10.5 - \frac{25}{6} = \frac{63-25}{6} = \frac{38}{6} \ cm$.For the objective,$f_0 = 1.9 \ cm$ and $v_0 = \frac{38}{6} \ cm$. Using the lens formula $\frac{1}{v_0} - \frac{1}{u_0} = \frac{1}{f_0}$:$\frac{1}{38/6} - \frac{1}{u_0} = \frac{1}{1.9} \Rightarrow \frac{6}{38} - \frac{1}{u_0} = \frac{10}{19} \Rightarrow \frac{3}{19} - \frac{10}{19} = \frac{1}{u_0}$.$\frac{1}{u_0} = -\frac{7}{19} \Rightarrow u_0 = -\frac{19}{7} \approx -2.71 \ cm$.Thus,the object should be placed at a distance of $2.7 \ cm$ from the objective.

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