A short straight object of height 100 cm lies before the central axis of a spherical mirror whose focal length has absolute value $$|f| = 40$$ cm. The image of object produced by the mirror is of height 25 cm and has the same orientation of the object. One may conclude from the information:

We are given: object height $$h_o = 100$$ cm, image height $$h_i = 25$$ cm, the image has the same orientation as the object, and the focal length has absolute value $$|f| = 40$$ cm.

The magnification is $$m = \frac{h_i}{h_o} = \frac{25}{100} = \frac{1}{4}$$. Since the image has the same orientation as the object (erect image), the magnification is positive: $$m = +\frac{1}{4}$$.

For mirrors, the magnification formula is $$m = -\frac{v}{u}$$. A positive magnification with magnitude less than 1 indicates a virtual, erect, and diminished image. This is the characteristic behaviour of a convex mirror for a real object.

For a convex mirror (using the sign convention where distances measured in the direction of incident light are positive), the focal length is $$f = +40$$ cm, and the object distance is negative ($$u < 0$$) since the object is in front of the mirror.

From $$m = -\frac{v}{u} = +\frac{1}{4}$$, we get $$v = -\frac{u}{4}$$. Since $$u < 0$$, we have $$v > 0$$, meaning the image is formed behind the mirror (virtual image).

Substituting into the mirror equation $$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$: $$\frac{1}{-u/4} + \frac{1}{u} = \frac{1}{40}$$, which gives $$-\frac{4}{u} + \frac{1}{u} = \frac{1}{40}$$, so $$-\frac{3}{u} = \frac{1}{40}$$, yielding $$u = -120$$ cm.

Then $$v = -\frac{(-120)}{4} = +30$$ cm. The positive value of $$v$$ confirms that the image is virtual and located behind the convex mirror, i.e., on the opposite side of the mirror from the object.

We can verify: $$\frac{1}{30} + \frac{1}{-120} = \frac{4 - 1}{120} = \frac{3}{120} = \frac{1}{40} = \frac{1}{f}$$. This confirms our answer.

Hence the correct answer is Option 4: Image is virtual, opposite side of convex mirror.