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(A) For a symmetric biconvex lens with radii $R$,the focal length $f$ is given by $1/f = (n-1)(2/R)$.When cut vertically,the resulting plano-convex le

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increased two times

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(A) For a symmetric biconvex lens with radii $R$,the focal length $f$ is given by $1/f = (n-1)(2/R)$.When cut vertically,the resulting plano-convex lens has a focal length $f'$ such that $1/f' = (n-1)(1/R) = 1/(2f)$,which implies $f' = 2f$.The magnification of a compound microscope is given by $m = (L/f_o) 	imes (D/f_e)$,where $L$ is the tube length,$f_o$ is the focal length of the objective,and $f_e$ is the focal length of the eyepiece.To keep the magnification $m$ constant while the object distance $u_o$ is unchanged,the image distance $v_o$ must remain constant. However,the problem specifies that the magnification $m$ must be retained. Since $m \propto L/f_o$,if $f_o$ is replaced by $f' = 2f_o$,then to keep $m$ constant,the tube length $L$ must be increased by a factor of $2$ (i.e.,$L' = 2L$).

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