A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that the end closer to the pole is 20 cm away from it. Find the length of the image.

The position of the image of the nearer end to the mirror can be found out using the mirror equation, 1/v + 1/u = 1/f ⟹ 1/v1 = 1/10 - 1/20 ⟹ v1 = 20 cm. Similarly, the position of the image of the farther end to the mirror can be found out as 1/v2 = 1/(-10) - 1/(20+10) ⟹ v2 = -15 cm. Hence the length of the image = 20 cm - (-15 cm) = 35 cm. There appears to be an error in the original solution provided. The calculation should be: 1/v + 1/u = 1/f. For the closer end: 1/v1 + 1/20 = 1/10; v1 = 20cm. For the farther end: 1/v2 + 1/30 = 1/10; v2 = 15cm. Length of the image = 20 - 15 = 5cm.