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 ______.

A concave mirror has focal length $$f = 20\text{ cm}$$. A thin rod of length $$10\text{ cm}$$ is placed along the principal axis so that its midpoint is at $$40\text{ cm}$$ from the pole of the mirror.

Let the two ends of the rod be at object distances $$u_1$$ and $$u_2$$ from the mirror. Then

$$u_1 = 40 - 5 = 35\text{ cm},\quad u_2 = 40 + 5 = 45\text{ cm}$$

We use the mirror formula:

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\quad-(1)$$

For the end at $$u_1 = 35\text{ cm}$$, substitute into $$(1)$$:

$$\frac{1}{v_1} = \frac{1}{20} - \frac{1}{35} = \frac{7 - 4}{140} = \frac{3}{140}$$

Thus,

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

For the end at $$u_2 = 45\text{ cm}$$, substitute into $$(1)$$:

$$\frac{1}{v_2} = \frac{1}{20} - \frac{1}{45} = \frac{9 - 4}{180} = \frac{5}{180} = \frac{1}{36}$$

Thus,

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

Both images lie on the same side of the mirror, so the image of the rod extends from $$v_2$$ to $$v_1$$. Hence the image length is

$$\text{Image length} = \bigl|v_1 - v_2\bigr| = \frac{140}{3} - 36 = \frac{140 - 108}{3} = \frac{32}{3}\text{ cm}$$

Since the image length is $$\frac{x}{3}\text{ cm}$$, we have

$$\frac{x}{3} = \frac{32}{3}\quad\Longrightarrow\quad x = 32$$