Animated Solution for Physics - Optics: A short linear object of length $b$ lies along the axis of a concave mirror of focal length $f$ at a distance $u$ from the pole of the mirror. The size of the image is approximately equal to

A Tiny Object on the Axis

Imagine you are setting up an optics experiment. You have a concave mirror, and instead of placing an object upright like a candle, you lay a tiny pin flat along the principal axis. This pin has a very short length, which we will call $b$. It is placed at a distance $u$ from the pole of the mirror. Our mission is to find the length of the image formed by this mirror.

Because the object is extremely short, we can treat its length as an infinitesimally small change in the object distance. Let's denote this tiny length as $du$. Consequently, the length of the image will be the corresponding tiny change in the image distance, which we will call $dv$. To bridge the gap between the object and the image, we rely on the fundamental mirror formula:

To find how a small change in $u$ affects $v$, we can apply a little calculus. By differentiating the entire mirror formula with respect to $u$, we can find the exact relationship between $dv$ and $du$. Remember, for a given mirror, the focal length $f$ is a constant, so its derivative is zero.

Differentiating both sides gives us:

Let's rearrange this equation to solve for $dv$, which represents the size of our image:

The negative sign here is physically significant—it tells us that the image is longitudinally inverted. If the head of the pin is further away from the mirror, the head of the image will be closer. However, we are only interested in the actual size (magnitude) of the image. Since the object length is $|du|=b$, the image length is:

We have a beautiful expression for the image size, but there is a catch. Our final answer must be in terms of the given variables: $b$, $u$, and $f$. The image distance $v$ is an unknown that we must eliminate. We need to express the ratio $\frac{v}{u}$ entirely in terms of $u$ and $f$.

Let's return to the mirror formula and solve for $\frac{1}{v}$:

Flipping both sides gives us $v$:

Now, dividing by $u$, we find our required ratio:

Finally, we substitute this ratio back into our equation for the image size:

And there we have it! The size of the image is $b$ times the square of $\frac{f}{u-f}$. This concept is known as longitudinal magnification. For infinitesimally short objects, the longitudinal magnification is always equal to the negative square of the transverse magnification ($m_L=-m^2$). This is a powerful shortcut to keep in your arsenal for competitive exams!