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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} = \fr

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$2.7$

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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 intermediate image 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 lens, using the lens formula $\frac{1}{v_0} - \frac{1}{u_0} = \frac{1}{f_0}$ with $f_0 = 1.9 \,cm$:$\frac{6}{38} - \frac{1}{u_0} = \frac{1}{1.9} \Rightarrow \frac{1}{u_0} = \frac{6}{38} - \frac{10}{19} = \frac{6 - 20}{38} = -\frac{14}{38}$.$u_0 = -\frac{38}{14} \approx -2.71 \,cm$.Thus, the object should be placed at a distance of $2.7 \,cm$ from the objective.

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