A mirror is used to produce an image with magnification of $\frac{1}{4}$. If the distance between object and its image is 40 cm, then the focal length of the mirror is ________.

<p>Magnification for a mirror is </p> <p>$m=-\frac{v}{u}$</p> <p>Given $m=\frac{1}{4}$, so </p> <p>$-\frac{v}{u}=\frac{1}{4}\;\Rightarrow\; v=-\frac{u}{4}$</p> <p>For a real object, $u<0$. Since $m$ is positive, the image is virtual, so $v>0$. </p> <p>Let $u=-x$ (where $x>0$). Then </p> <p>$v=-\frac{-x}{4}=\frac{x}{4}$</p> <p>Distance between object and image (object in front, image behind) is </p> <p>$x+v=40$</p> <p>$x+\frac{x}{4}=40$</p> <p>$\frac{5x}{4}=40 \Rightarrow x=32\ \text{cm}$</p> <p>So,</p> <p>$u=-32\ \text{cm},\quad v=+8\ \text{cm}$</p> <p>Now use mirror formula:</p> <p>$$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}=\frac{1}{8}-\frac{1}{32}=\frac{4-1}{32}=\frac{3}{32}$$</p> <p>$f=\frac{32}{3}\approx 10.7\ \text{cm}$</p> <p><strong>Correct option: C) $10.7\ \text{cm}$</strong></p>