Given values are object distance $u = -9$ cm, focal length $f = 9$ cm, and area of each square $A = 1 \text{ mm}^2$.

Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$, we substitute the values: $\frac{1}{v} - \frac{1}{-9} = \frac{1}{9}$. This simplifies to $\frac{1}{v} = 0$, so the image distance $v = \infty$.

The magnification produced by the lens is $m = \frac{v}{u} = \frac{\infty}{-9} = \infty$. Since the linear magnification is infinite, the area of each square in the virtual image is also infinite.

The angular magnification (magnifying power) of the lens when the image is formed at infinity is given by $M = \frac{D}{f}$. Taking the least distance of distinct vision $D = 25$ cm, we get $M = \frac{25}{9} \approx 2.8$.

No, the magnification in (a) is not equal to the magnifying power in (b). The magnification in (a) is infinite, while the magnifying power in (b) is 2.8. Magnification refers to the ratio of the absolute size of the image to the object, whereas magnifying power is the ratio of the angle subtended by the image at the eye to the angle subtended by the object at the near point.