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(B) Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.For a concave mirror,the focal length $f$ is taken as negative,so $f_{mirror}

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$-3$

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(B) Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.For a concave mirror,the focal length $f$ is taken as negative,so $f_{mirror} = -f$.The object distance $u$ is also negative,so $u = -1.5 f = -\frac{3 f}{2}$.Substituting these values into the mirror formula:$\frac{1}{v} + \frac{1}{-1.5 f} = \frac{1}{-f}$$\frac{1}{v} - \frac{2}{3 f} = -\frac{1}{f}$$\frac{1}{v} = -\frac{1}{f} + \frac{2}{3 f}$$\frac{1}{v} = \frac{-3 + 2}{3 f}$$\frac{1}{v} = -\frac{1}{3 f}$Therefore,$v = -3 f$. The image is formed at a distance of $3 f$ in front of the mirror.

An object is placed on the principal axis of a concave mirror at a distance of $1.5 f$ ($f$ is the focal length). The image will be at ....... $f$.

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