When an object is placed at a distance of 25 | vedclass

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(D) For a concave mirror,the magnification $m$ is given by $m = \frac{f}{f-u}$.For the first position,$u_1 = -25 \ cm$,so $m_1 = \frac{f}{f - (-25)} =

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$20$

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(D) For a concave mirror,the magnification $m$ is given by $m = \frac{f}{f-u}$.For the first position,$u_1 = -25 \ cm$,so $m_1 = \frac{f}{f - (-25)} = \frac{f}{f+25}$.For the second position,the object is moved $15 \ cm$ farther,so $u_2 = -(25 + 15) = -40 \ cm$. Thus,$m_2 = \frac{f}{f - (-40)} = \frac{f}{f+40}$.Given $\frac{m_1}{m_2} = 4$,we substitute the expressions: $\frac{f/(f+25)}{f/(f+40)} = 4$.This simplifies to $\frac{f+40}{f+25} = 4$.$f + 40 = 4(f + 25) \implies f + 40 = 4f + 100$.$3f = -60 \implies f = -20 \ cm$.The focal length of a concave mirror is $20 \ cm$ (magnitude).

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