Animated Solution for Physics - Optics: 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 $\frac{\pi }{\text{n}}$ rad, where n is _________.

Imagine you are tasked with finding the image of a rod placed in front of a lens. If the rod is perfectly vertical, it's a straightforward transverse magnification problem. If it's perfectly horizontal, it's a longitudinal magnification problem. But what happens when the rod is tilted at an angle? This is where many students stumble, but the solution is surprisingly elegant if we break it down into fundamental steps.

The most robust strategy for dealing with extended, tilted objects is to stop looking at the rod as a single entity. Instead, treat it as a collection of points. To find the position and orientation of the image rod, we only need to find the images of its two extreme endpoints: the top end and the bottom end.

Let's start with the bottom end, which conveniently lies right on the principal axis. The problem states the rod intersects the axis at a distance of $\frac{40}{3}$ cm from the lens. Using the standard lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$, we substitute $u_2=-\frac{40}{3}$ cm and $f=10$ cm. A quick calculation reveals that the image of this bottom end forms at $v_2=40$ cm on the other side of the lens.

Now for the slightly trickier part: the top end. The rod has a length of $2$ cm and is tilted at an angle of $\frac{2\pi }{3}$ radians (or $120^{\circ }$) with the positive principal axis. This means it makes a $60^{\circ }$ angle with the negative x-axis.

Using basic trigonometry, we can find the exact $(x,y)$ coordinates of this top end. The x-coordinate is shifted from the bottom end by the horizontal component of the rod's length: $x_1=-\frac{40}{3}-2\cos (60^{\circ })=-\frac{43}{3}$ cm. The y-coordinate is simply the vertical component: $y_1=2\sin (60^{\circ })=\sqrt{3}$ cm.

With the object distance of the top end ($u_1=-\frac{43}{3}$ cm) known, we apply the lens formula once again. This gives us the image position $v_1=\frac{430}{13}$ cm. To find how high this image point is, we use the transverse magnification formula $m=\frac{v}{u}$. The magnification comes out to be $-\frac{30}{13}$, meaning the image is inverted. Multiplying this by the object height $\sqrt{3}$ cm gives us the image height $h_i=-\frac{30\sqrt{3}}{13}$ cm.

We now have the coordinates of both ends of the image rod! The bottom end is at $(40,0)$ and the top end is at $(\frac{430}{13},-\frac{30\sqrt{3}}{13})$. The angle $\alpha$ that the image rod makes with the principal axis is directly related to the slope of the line connecting these two points.

By calculating $\tan \alpha =\frac{|h_i|}{v_2-v_1}$, we get $\frac{1}{\sqrt{3}}$. This beautifully simplifies to $\alpha =30^{\circ }$, or $\frac{\pi }{6}$ radians. Comparing this to the given expression $\frac{\pi }{n}$, we confidently conclude that $n=6$.

Always remember, when faced with a complex extended object, breaking it down into its constituent points is your most powerful tool!