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(D) Initially,the object distance $u_o = -3\,cm$ and the focal length of the objective $f_o = 2\,cm$. The image distance $v_o$ formed by the objective

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$9$

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(D) Initially,the object distance $u_o = -3\,cm$ and the focal length of the objective $f_o = 2\,cm$. The image distance $v_o$ formed by the objective is given by the lens formula $\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o}$.$v_o = \frac{u_o f_o}{u_o + f_o} = \frac{(-3)(2)}{-3 + 2} = \frac{-6}{-1} = 6\,cm$.When a lens of focal length $f_1 = 10\,cm$ is removed from the objective system,the new focal length $f'_o$ is given by $\frac{1}{f'_o} = \frac{1}{f_o} - \frac{1}{f_1}$.$\frac{1}{f'_o} = \frac{1}{2} - \frac{1}{10} = \frac{5-1}{10} = \frac{4}{10} = \frac{2}{5}$. Thus,$f'_o = 2.5\,cm$.The new image distance $v'_o$ for the same object distance $u_o = -3\,cm$ is $v'_o = \frac{u_o f'_o}{u_o + f'_o} = \frac{(-3)(2.5)}{-3 + 2.5} = \frac{-7.5}{-0.5} = 15\,cm$.To refocus the image,the eyepiece must be moved by the shift in the position of the image formed by the objective,which is $\Delta v = |v'_o - v_o| = |15 - 6| = 9\,cm$.

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