Animated Solution for Physics - Optics: The magnifying power of a telescope with tube length $60\text{ cm}$ is $5$. What is the focal length of its eyepiece?

Since the days of Galileo, telescopes have been our primary tool for unlocking the mysteries of the cosmos. By cleverly combining lenses, we can gather light from distant stars and magnify them so they appear right before our eyes. But how exactly do these lenses work together? The secret lies in two crucial parameters: the tube length and the magnifying power.

In this problem, we are given a telescope with a tube length of $60\text{ cm}$ and a magnifying power of $5$. Our mission is to find the focal length of its eyepiece. Let's break down the physics behind this setup.

When you look through a telescope for a long time, you want your eyes to be as relaxed as possible. The human eye is most relaxed when it is looking at an object infinitely far away. Therefore, telescopes are usually set up in what we call normal adjustment.

In normal adjustment, the final image produced by the telescope is formed at infinity. For this to happen, the image formed by the first lens (the objective) must fall exactly on the focal point of the second lens (the eyepiece).

Because the object (a star or planet) is already at infinity, the objective lens forms its image at its own focal point, $f_o$. Since this image must also sit at the focal point of the eyepiece, $f_e$, it means that the two focal points must perfectly coincide!

Because the focal points of the two lenses meet at the exact same spot inside the tube, the total distance between the two lenses—which is the tube length ($L$)—is simply the sum of their focal lengths.

We are given that the tube length is $60\text{ cm}$. So, our first equation is:

The primary job of a telescope is to make things look bigger. The magnifying power ($m$) in normal adjustment is defined as the ratio of the angle subtended by the image at the eye to the angle subtended by the object at the unaided eye. Geometrically, this simplifies beautifully to the ratio of the focal lengths:

We are told the magnifying power is $5$. This gives us our second equation:

We now have a straightforward system of two linear equations. Let's solve them!

From the magnification equation, we can express the focal length of the objective in terms of the eyepiece by cross-multiplying:

Now, we substitute this relationship back into our tube length equation. We replace $f_o$ with $5f_e$:

Combining the terms gives us:

Finally, dividing both sides by $6$, we find the focal length of the eyepiece:

And there we have it! The focal length of the eyepiece is exactly $10\text{ cm}$.

Notice how the objective lens has a focal length of $50\text{ cm}$ ($5\times 10\text{ cm}$), which is much larger than the eyepiece. This is a hallmark of astronomical telescopes: a large objective focal length provides high magnification and allows for a larger aperture to gather more starlight. The math perfectly reflects the physical design of the instrument!