A rod of length 2 cm makes an angle $$\frac{2\pi}{3}$$ rad with the principal axis of a thin convex lens. The lens has a focal length of 10 cm and is placed at a distance of $$\frac{40}{3}$$ cm from the object as shown in the figure. The height of the image is $$\frac{30\sqrt{3}}{13}$$ cm and the angle made by it with respect to the principal axis is $$\alpha$$ rad. The value of $$\alpha$$ is $$\frac{\pi}{n}$$ rad, where $$n$$ is ______.

Let the convex lens be taken as the origin and its principal axis as the positive $$x$$-axis. Distances measured to the left of the lens are negative.

The focal length is $$f = 10\text{ cm}$$ and the foot of the rod on the axis is kept at an object distance
$$u_A = -\frac{40}{3}\text{ cm}$$.

Object end A (on the principal axis)
Applying the thin-lens formula $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$,

$$\frac{1}{v_A} - \frac{1}{-\frac{40}{3}} = \frac{1}{10} \\ \Rightarrow \frac{1}{v_A} + \frac{3}{40} = \frac{1}{10} \\ \Rightarrow \frac{1}{v_A} = \frac{1}{40} \\ \Rightarrow v_A = 40\text{ cm}$$

The transverse magnification for this point is
$$m_A = \frac{v_A}{u_A} = \frac{40}{-\frac{40}{3}} = -3$$.
Since $$y_A = 0$$, the image coordinate is $$y'_A = m_A y_A = 0$$.

Second end B of the rod
The rod has length $$L = 2\text{ cm}$$ and is inclined at
$$\theta = \frac{2\pi}{3} = 120^{\circ}$$ to the axis.
Its components are $$\Delta x = L\cos\theta = 2(-\tfrac12) = -1\text{ cm},\; \Delta y = L\sin\theta = 2\!\left(\tfrac{\sqrt3}{2}\right) = \sqrt3\text{ cm}$$.

Hence the coordinates of B are
$$x_B = u_A+\Delta x = -\frac{40}{3}-1 = -\frac{43}{3}\text{ cm},\; y_B = 0+\Delta y = \sqrt3\text{ cm}$$.

Using the lens formula again for B,

$$\frac{1}{v_B} - \frac{1}{-\frac{43}{3}} = \frac{1}{10} \\ \Rightarrow \frac{1}{v_B} + \frac{3}{43} = \frac{1}{10} \\ \Rightarrow \frac{1}{v_B} = \frac{13}{430} \\ \Rightarrow v_B = \frac{430}{13}\text{ cm}$$

The magnification for B is
$$m_B = \frac{v_B}{u_B} = \frac{\tfrac{430}{13}}{-\tfrac{43}{3}} = -\frac{30}{13}$$.

Therefore the image coordinate of B is
$$y'_B = m_B y_B = -\frac{30}{13}\,\sqrt3\text{ cm}$$,
while $$x'_B = v_B = \frac{430}{13}\text{ cm}$$.

Length and orientation of the image rod
The differences between image coordinates are
$$\Delta y' = y'_B - y'_A = -\frac{30\sqrt3}{13}\text{ cm} \\ \Delta x' = x'_B - x'_A = \frac{430}{13} - 40 = \frac{430 - 520}{13} = -\frac{90}{13}\text{ cm}$$.

The magnitude of the height is indeed
$$|\Delta y'| = \frac{30\sqrt3}{13}\text{ cm},$$
matching the data given.

The angle $$\alpha$$ that the image makes with the axis is obtained from

$$\tan\alpha = \left|\frac{\Delta y'}{\Delta x'}\right| = \frac{30\sqrt3/13}{90/13} = \frac{\sqrt3}{3} = \tan\!\left(\frac{\pi}{6}\right).$$

Hence $$\alpha = \frac{\pi}{6}\text{ rad}$$. Writing $$\alpha = \frac{\pi}{n}$$ gives $$n = 6$$.

Answer (numerical): 6