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(A) The magnification formula for a spherical mirror is $m = \frac{f}{f-u}$.Given $m = \frac{1}{2}$ and $u = -40\ cm$,we have $\frac{1}{2} = \frac{f}{

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$40\ cm$ away from the mirror.

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(A) The magnification formula for a spherical mirror is $m = \frac{f}{f-u}$.Given $m = \frac{1}{2}$ and $u = -40\ cm$,we have $\frac{1}{2} = \frac{f}{f - (-40)}$.$f + 40 = 2f$,which gives $f = 40\ cm$.Now,to obtain a magnification of $m = \frac{1}{3}$,we use the same formula: $\frac{1}{3} = \frac{40}{40 - u'}$.$40 - u' = 120$,so $u' = -80\ cm$.The object was initially at $40\ cm$ and is now at $80\ cm$ from the mirror.Therefore,the object must be moved $80 - 40 = 40\ cm$ away from the mirror.

When an object is placed $40\ cm$ away from a spherical mirror,an image of magnification $\frac{1}{2}$ is produced. To obtain an image with magnification of $\frac{1}{3}$,the object is to be moved $:$

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