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(D) For a spherical mirror,the magnification $m$ is given by $m = \frac{f}{f - u}$.In the first case,the object distance $u_1 = -25 \; cm$. Thus,$m_1

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$-20$

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(D) For a spherical mirror,the magnification $m$ is given by $m = \frac{f}{f - u}$.In the first case,the object distance $u_1 = -25 \; cm$. Thus,$m_1 = \frac{f}{f - (-25)} = \frac{f}{f + 25}$.In the second case,the object is moved $15 \; cm$ further away,so the new object distance $u_2 = -(25 + 15) = -40 \; cm$. Thus,$m_2 = \frac{f}{f - (-40)} = \frac{f}{f + 40}$.Given the ratio $\frac{m_1}{m_2} = 4$,we have:$\frac{f / (f + 25)}{f / (f + 40)} = 4$$\frac{f + 40}{f + 25} = 4$$f + 40 = 4(f + 25)$$f + 40 = 4f + 100$$3f = -60$$f = -20 \; cm$.

When an object is placed at a distance of $25 \; cm$ from a mirror,the magnification is $m_1$. The object is moved $15 \; cm$ 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: (in $; cm$)

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