A convex mirror of focal length f forms an im | vedclass

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(A) For a convex mirror,the magnification $m$ is given by $m = +\frac{1}{n}$ because the image formed is virtual and erect.Using the magnification for

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$(n-1)f$

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(A) For a convex mirror,the magnification $m$ is given by $m = +\frac{1}{n}$ because the image formed is virtual and erect.Using the magnification formula $m = -\frac{v}{u}$,we have $\frac{1}{n} = -\frac{v}{u}$,which implies $v = -\frac{u}{n}$.According to the mirror formula,$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.Substituting the value of $v$,we get $\frac{1}{f} = \frac{1}{-u/n} + \frac{1}{u} = -\frac{n}{u} + \frac{1}{u} = \frac{1-n}{u}$.Rearranging for $u$,we get $u = f(1-n) = -(n-1)f$.Since the distance is the magnitude of $u$,the distance of the object from the mirror is $(n-1)f$.

$A$ convex mirror of focal length $f$ forms an image which is $\frac{1}{n}$ times the object. The distance of the object from the mirror is

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