Animated Solution for Physics - Optics: An astronomical telescope has an angular magnification of magnitude $5$ for far objects. The separation between the objective and the eyepiece is $36\text{ cm}$ and the final image is formed at infinity. The focal length $f_o$ of the objective and the focal length $f_e$ of the eyepiece are

Have you ever looked up at the night sky and wondered how a telescope brings distant galaxies right to your eye? The astronomical telescope is a marvel of optical engineering, yet its fundamental principles are beautifully simple. In this article, we will dissect a classic problem that explores the relationship between the focal lengths of the lenses, the length of the telescope tube, and its magnifying power.

Imagine you are looking through an astronomical telescope. It consists of two converging lenses: the objective lens, which faces the distant object, and the eyepiece, which you look through.

When we use a telescope to view far-off objects like stars or planets, the incoming light rays are essentially parallel. The objective lens takes these parallel rays and converges them to form a real, inverted image at its focal point.

Now, for the most relaxed viewing experience, we want the final image to be formed at infinity. This is known as normal adjustment. For this to happen, the image formed by the objective lens must lie exactly at the focal point of the eyepiece. When an object is at the focal point of a converging lens, the emerging rays are parallel, meaning the final image is at infinity.

Because the focal points of both lenses coincide, the total distance between the objective and the eyepiece—often called the tube length $L$—is simply the sum of their focal lengths:

In our specific problem, we are given that the separation between the lenses is $36\text{ cm}$. Therefore, our first master equation is:

The primary purpose of a telescope is to make distant objects appear larger, or more precisely, to increase the angle they subtend at our eye. This is measured by the angular magnification or magnifying power $M$.

For an astronomical telescope in normal adjustment, the angular magnification is the ratio of the angle subtended by the image to the angle subtended by the object. Through some elegant geometry, this ratio simplifies to the ratio of the focal lengths of the two lenses:

The problem states that the angular magnification has a magnitude of $5$. This gives us our second crucial equation:

This tells us a fundamental truth about telescope design: to get high magnification, you need an objective lens with a long focal length and an eyepiece with a short focal length.

Now that we have our two equations, the physics is complete, and we enter the realm of simple algebra. From the magnification equation, we can express the focal length of the objective in terms of the eyepiece:

Let's substitute this relationship back into our tube length equation:

Combining the terms, we get:

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

With the eyepiece focal length known, finding the objective's focal length is a breeze. We just multiply by $5$:

And there we have it! The focal length of the objective is $30\text{ cm}$, and the focal length of the eyepiece is $6\text{ cm}$. This perfectly matches option (d).

This problem beautifully illustrates the trade-offs in telescope design. If you want more magnification, you must either increase the focal length of the objective (which makes the telescope longer and bulkier) or decrease the focal length of the eyepiece. Understanding these simple linear relationships is the first step toward mastering optical instruments. Keep visualizing the rays, and the math will always follow!