Animated Solution for Physics - Optics: A thin rod of length $\frac{f}{3}$ is placed along the optic axis of a concave mirror of focal length $f$ such that its image which is real and elongated, just touches the rod. The magnification is ......

Imagine you are setting up an optical experiment. You have a concave mirror, and you place a thin rod along its principal axis. The problem presents a fascinating scenario: the image of the rod is real, elongated, and it perfectly touches the rod itself!

This isn't just a random arrangement; it's a beautifully constrained geometric puzzle. Let's break down the physics behind it.

For a real image to physically touch the object, there must be a point on the principal axis where the object and its image share the exact same coordinate.

Do you remember which point in front of a concave mirror has this magical property? Yes, it's the center of curvature ($C$). When an object is placed at $C$, its real, inverted image is formed exactly at $C$. Therefore, one end of our rod must be anchored at the center of curvature.

Now, the rod has a given length of $\frac{f}{3}$. Which way does it extend? Towards the mirror or away from it?

The problem states that the image is elongated (magnified). If we placed the rod beyond $C$, its image would form between $C$ and the focus ($F$), and it would be diminished. To get an elongated image, the object must be placed closer to the focus, meaning the rod extends from $C$ towards $F$.

Let's define the coordinates. The pole of the mirror is at the origin ($0,0$). The center of curvature is at a distance of $2f$ from the pole, so its coordinate is $u_1=-2f$.

Since the rod extends towards the mirror by a length of $\frac{f}{3}$, the coordinate of the other end, let's call it $P$, will be:

We know the image of the end at $C$ forms at $C$ ($v_1=-2f$). Now, we need to find where the image of end $P$ forms. We bring in our trusty mirror formula:

Substituting our values for point $P$:

Let's solve this equation for $v_2$. Moving the object term to the right side:

Taking the common denominator:

Inverting this gives us the position of the image of end $P$:

So, the image of the rod starts at $C$ ($-2f$) and extends to $-2.5f$. The length of this image rod is simply the difference between these two coordinates:

Finally, we need to find the longitudinal magnification. It is defined as the negative ratio of the image length to the object length. The negative sign accounts for the fact that the image is inverted relative to the object along the axis.

And there we have it! The magnification of the rod is $-1.5$. The beauty of this problem lies in how a simple physical constraint—the image touching the object—locks the entire geometry into place.