The relation between the linear magnification | vedclass

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Question

(B) The linear magnification $m$ for a spherical mirror is given by $m = -\frac{v}{u}$.From the mirror formula,we have $\frac{1}{f} = \frac{1}{v} + \f

Vedclass · Accepted Answer

$m = \frac{f}{f - u}$

Vedclass · Answer

(B) The linear magnification $m$ for a spherical mirror is given by $m = -\frac{v}{u}$.From the mirror formula,we have $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.Multiplying both sides by $u$,we get $\frac{u}{f} = \frac{u}{v} + 1$.Rearranging the terms,we get $\frac{u}{v} = \frac{u}{f} - 1 = \frac{u - f}{f}$.Taking the reciprocal,we get $\frac{v}{u} = \frac{f}{u - f}$.Since $m = -\frac{v}{u}$,we have $m = -\frac{f}{u - f} = \frac{f}{f - u}$.Therefore,the correct relation is $m = \frac{f}{f - u}$.

The relation between the linear magnification $m$,the object distance $u$,and the focal length $f$ is:

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