Two objects $$A$$ and $$B$$ are placed at 15 cm and 25 cm from the pole in front of a concave mirror having radius of curvature 40 cm. The distance between images formed by the mirror is:

We have a concave mirror with radius of curvature $$R = 40$$ cm, so focal length $$f = 20$$ cm. Objects $$A$$ and $$B$$ are placed at 15 cm and 25 cm from the pole.

Using $$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$ with $$f = -20$$ cm (concave mirror), for object $$A$$ at $$u = -15$$ cm we have $$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-20} - \frac{1}{-15} = -\frac{1}{20} + \frac{1}{15} = \frac{-3 + 4}{60} = \frac{1}{60}$$, which yields $$v_A = +60$$ cm (positive sign means the image is behind the mirror — virtual image).

Similarly, for object $$B$$ at $$u = -25$$ cm, $$\frac{1}{v} = \frac{1}{-20} - \frac{1}{-25} = -\frac{1}{20} + \frac{1}{25} = \frac{-5 + 4}{100} = -\frac{1}{100}$$ and hence $$v_B = -100$$ cm (negative sign means the image is in front of the mirror — real image).

Image $$A$$ is 60 cm behind the mirror and image $$B$$ is 100 cm in front of the mirror, so the distance between them is $$|v_A| + |v_B| = 60 + 100 = 160$$ cm. The answer is Option C: 160 cm.