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f-R Relationship

When an objec...

Question

When an object is placed at a distance of $25 c m$ from a mirror, the magnification is $m_{1}$ . The object is moved $15 c m$ away with respect to the earlier position, magnification becomes $m_{2}$ , If $\frac{m_{1}}{m_{2}} = 4$ , the focal length of the mirror is

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Solution

$u_{1} = - 25 c m, u_{2} = - (25 + 15) = - 40 c m$
$m_{1} = - \frac{v_{1}}{u_{1}}$
$m_{2} = - \frac{v_{2}}{u_{2}}$
$\frac{m_{1}}{m_{2}} = \frac{v_{1}}{v_{2}} \times \frac{u_{2}}{u_{1}}$
$4 = \frac{v_{1}}{v_{2}} . \frac{40}{25}$
$v_{1} = 2.5 v_{2}$
$\frac{1}{u_{1}} + \frac{1}{v_{1}} = \frac{1}{f} \Rightarrow \frac{1}{- 25} + \frac{1}{2.5 v_{2}} = \frac{1}{f}$ ....(1)
$\frac{1}{u_{2}} + \frac{1}{v_{2}} = \frac{1}{f} \Rightarrow \frac{1}{- 40} + \frac{1}{v_{2}} = \frac{1}{f}$ ....(2)
Divinding (2) by 2.5 , $\frac{1}{- 100} + \frac{1}{2.5 v_{2}} = \frac{1}{f}$ ......(3)
$(1) - (3) \Rightarrow \frac{1}{- 25} + \frac{1}{100} = \frac{1}{f} - \frac{1}{2.5 f}$
$\frac{- 3}{100} = \frac{1.5}{2.5 f}$
$f = - 20 c m$

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