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

Question

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Answer 1

From the formula, $v = \frac{u f}{u - f}$

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

From the question,

For first case, object distance $u_{1} = - 25 c m$

After the object is moved 15 cm away from the mirror, new object distance becomes $u_{2} = - 40 c m$

$∴$ $\frac{m _{1}}{m _{2}} = \frac{f - u _{2}}{f - u 1} = 4$

Or $\frac{f + 4 0}{f + 2 5} = 4$

Or $3 f = - 60 c m$

$⟹ f = - 20 c m$

Answer 2

From the formula,

v = \frac{u f}{u - f}

Magnification

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

From the question,

For first case, object distance

u_{1} = - 25 c m

After the object is moved 15 cm away from the mirror, new object distance becomes

u_{2} = - 40 c m

∴

\frac{m _{1}}{m _{2}} = \frac{f - u _{2}}{f - u 1} = 4

Or

\frac{f + 4 0}{f + 2 5} = 4

Or

3 f = - 60 c m

⟹ f = - 20 c m