A thin cylindrical rod of length 10 cm is placed horizontally on the principle axis of a concave mirror of focal length 20 cm. The rod is placed in a such a way that mid point of the rod is at 40 cm from the pole of mirror. The length of the image formed by the mirror will be $$\frac{x}{3}$$ cm. The value of $$x$$ is _____.

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \implies v = \frac{uf}{u - f}$$

Focal length of concave mirror $$f = -20 \text{ cm}$$

Length of rod $$L = 10 \text{ cm}$$

Midpoint of rod is at $$u_{\text{mid}} = -40 \text{ cm}$$

Since the midpoint is at $$40 \text{ cm}$$ from the pole, the rod extends $$5 \text{ cm}$$ on either side.

Position of the closer end ($$A$$): $$u_1 = -40 + 5 = -35 \text{ cm}$$

Position of the farther end ($$B$$): $$u_2 = -40 - 5 = -45 \text{ cm}$$

For end $$A$$ ($$u_1 = -35 \text{ cm}$$):

$$v_1 = \frac{(-35)(-20)}{-35 - (-20)}$$

$$v_1 = \frac{700}{-15}$$

$$v_1 = -\frac{140}{3} \text{ cm}$$

For end $$B$$ ($$u_2 = -45 \text{ cm}$$):

$$v_2 = \frac{(-45)(-20)}{-45 - (-20)}$$

$$v_2 = \frac{900}{-25}$$

$$v_2 = -36 \text{ cm}$$

The length of the image is the distance between these two points:

$$L' = |v_1 - v_2|$$

$$L' = \left| -\frac{140}{3} - (-36) \right|$$

$$L' = \left| -\frac{140}{3} + \frac{108}{3} \right|$$

$$L' = \left| -\frac{32}{3} \right| = \frac{32}{3} \text{ cm}$$