An object of height 3 cm is placed 15 cm from a concave mirror of focal length 10 cm. a) Calculate the magnification of the image. b) Calculate the height of the image. c) Is the image magnified or diminished? Explain your answer.

The mirror formula is $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$, where $f$ is the focal length, $u$ is the object distance, and $v$ is the image distance. For a concave mirror, $f$ is positive.
Given: $f = 10 \text{ cm}$, $u = 15 \text{ cm}$.

The magnification formula is $M = -\frac{v}{u}$.

The magnification of the image is $\boxed{-2}$.

Given: $h_o = 3 \text{ cm}$, $M = -2$.

The height of the image is $\boxed{-6 \text{ cm}}$. The negative sign indicates that the image is inverted.

The absolute value of the magnification is $|M| = |-2| = 2$. Since $|M| > 1$, the image is magnified.
Alternatively, the absolute height of the image is $|h_i| = |-6 \text{ cm}| = 6 \text{ cm}$, which is greater than the object height $h_o = 3 \text{ cm}$. Therefore, the image is magnified.