1. Interpret the object geometry

A rod of length $L=2\,\text{cm}$ is inclined at angle $\theta = \frac{2\pi}{3}=120^\circ$ with the principal axis.

So its vertical extent is $\Delta y = L\sin 120^\circ = 2\cdot \frac{\sqrt{3}}{2}=\sqrt{3}\,\text{cm}$ and its horizontal extent is $\Delta x = L\cos 120^\circ = 2\cdot \left(-\frac12\right)=-1\,\text{cm}.$

Thus, between the two ends of the rod, one end is $1\,$cm closer to the lens than the other.

2. Object distances of the two ends

The lens is at a distance $u_1=\frac{40}{3}\,\text{cm}$ from one end of the object.

Because $\Delta x=-1\,$cm, the other end is at distance $u_2=\frac{40}{3}+1=\frac{43}{3}\,\text{cm}.$

So the two ends are at object distances $u_1=\frac{40}{3},\qquad u_2=\frac{43}{3}.$

3. Find image distances using lens formula

For a thin convex lens, $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ with Cartesian signs, or equivalently for real object/real image magnitudes, $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}.$

Given $f=10\,$cm:

For end 1:

$\frac{1}{10}=\frac{1}{v_1}+\frac{3}{40}$ $\frac{1}{v_1}=\frac{1}{10}-\frac{3}{40}=\frac{1}{40}$ $v_1=40\,\text{cm}.$

For end 2:

$\frac{1}{10}=\frac{1}{v_2}+\frac{3}{43}$ \frac{1}{v_2}=\frac{1}{10}-\frac{3}{43}=rac{43-30}{430}=\frac{13}{430} $v_2=\frac{430}{13}\,\text{cm}.$

4. Find image heights of the two ends

Magnification for each point: $m=\frac{v}{u}$ (in magnitude; image is inverted, but for size difference we track coordinates carefully).

Take the first end on the principal axis, so its object height is $0$. The second end has height $y_2=\sqrt{3}\,\text{cm}.$

Then image height of second end is $y_2'=-\frac{v_2}{u_2}y_2$ with $\frac{v_2}{u_2}=\frac{430/13}{43/3}=\frac{430}{13}\cdot\frac{3}{43}=\frac{30}{13}.$

Hence $y_2'=-\frac{30\sqrt{3}}{13}\,\text{cm}.$

Its magnitude is exactly the given height, $\frac{30\sqrt{3}}{13}\,\text{cm},$ so the setup is consistent.

The first end lies on axis, so its image is also on axis: $y_1'=0$.

5. Find inclination of the image rod

The two image points are:

• first end at $(v_1,0)=(40,0)$,
• second end at $(v_2,-30\sqrt{3}/13)=\left(\frac{430}{13},-\frac{30\sqrt{3}}{13}\right)$.

So horizontal separation between their images is \Delta x' = v_2-v_1=\frac{430}{13}-40=rac{430-520}{13}=-\frac{90}{13}.

Vertical separation is $\Delta y'=-\frac{30\sqrt{3}}{13}.$

Therefore, $\tan \alpha = \left|\frac{\Delta y'}{\Delta x'}\right|=\frac{30\sqrt{3}/13}{90/13}=\frac{\sqrt{3}}{3}.$

So the acute angle the image makes with the principal axis is $\alpha=\frac{\pi}{6}.$

Since $\alpha=\frac{\pi}{n}$, $n=6.$

6. Compare with stored answer

Stored correct answer: $5.95$ to $6.05$

Our derived value is $n=6,$ which matches the stored answer range.