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(A) For a concave mirror,the focal length $f$ is taken as negative,so $f_{mirror} = -f$. The object distance $u$ is also negative,so $u = -4f$. The he

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$2$

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(A) For a concave mirror,the focal length $f$ is taken as negative,so $f_{mirror} = -f$. The object distance $u$ is also negative,so $u = -4f$. The height of the object $h_o = 6 \ cm$.Using the mirror formula,$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$,we get $\frac{1}{v} + \frac{1}{-4f} = \frac{1}{-f}$.$\frac{1}{v} = \frac{1}{4f} - \frac{1}{f} = \frac{1 - 4}{4f} = -\frac{3}{4f}$.Thus,$v = -\frac{4f}{3}$.The magnification $m$ is given by $m = -\frac{v}{u} = \frac{h_i}{h_o}$.$m = -\frac{(-4f/3)}{(-4f)} = -\frac{1}{3}$.Therefore,$h_i = m 	imes h_o = -\frac{1}{3} 	imes 6 \ cm = -2 \ cm$.The length (magnitude) of the image is $2 \ cm$.

An object of length $6 \ cm$ is placed on the principal axis of a concave mirror of focal length $f$ at a distance of $4f$. The length of the image will be......$cm$.

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