An astronomical telescope has an angular magnification of magnitude 5 for far objects. The separatio

Q: Why is the final image formed at infinity?

In normal adjustment, the telescope is set up so that the eye is relaxed. This happens when parallel rays enter the eye, meaning the final image is effectively at infinity.

Q: Why is the distance between the lenses exactly $f_o + f_e$?

For the final image to be at infinity, the intermediate image formed by the objective must lie exactly at the focal point of the eyepiece. Since it already lies at the focal point of the objective, the total distance is the sum of their focal lengths.

Q: Does the negative sign in magnification matter here?

The angular magnification is technically $-f_o/f_e$ because the final image is inverted relative to the original object. However, the problem specifies the *magnitude* is 5, so we use the absolute value.

Q: What if the final image was formed at the near point instead of infinity?

If the image was at the near point ($D$), the length of the telescope would be $f_o + \frac{f_e D}{f_e + D}$ and the magnification would be $\frac{f_o}{f_e}\left(1 + \frac{f_e}{D}\right)$.

Q: Why is the objective lens always larger than the eyepiece in an astronomical telescope?

A larger objective lens gathers more light, which is crucial for observing faint, distant astronomical objects. It also improves the resolving power of the telescope.

Question

An astronomical telescope has an angular magnification of magnitude $5$ for far objects. The separation between the objective and the eyepiece is $36 	ext{ cm}$ 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

Options:

(A) $f_o = 45 	ext{ cm}$ and $f_e = -9 	ext{ cm}$

(B) $f_o = 50 	ext{ cm}$ and $f_e = 10 	ext{ cm}$

(C) $f_o = 7.2 	ext{ cm}$ and $f_e = 5 	ext{ cm}$

(D) $f_o = 30 	ext{ cm}$ and $f_e = 6 	ext{ cm}$

Sigma Percentile · Accepted Answer

### Unlocking the Secrets of the Astronomical Telescope

Imagine you are gazing at the stars through an astronomical telescope. To view these distant celestial bodies comfortably, the telescope is often adjusted so that the final image forms at infinity. This state is known as **normal adjustment**, and it allows your eye muscles to remain completely relaxed while observing.

### The Setup: Normal Adjustment

In an astronomical telescope, there are two converging lenses: the **objective lens** (which faces the object) and the **eyepiece** (which you look through).

When light rays from a distant star enter the objective lens, they converge to form a real, inverted image at its focal plane. For the final image to form at infinity, this intermediate image must act as an object that lies exactly at the focal point of the eyepiece.

Because the intermediate image is at the focal point of both lenses simultaneously, the total distance $L$ between the two lenses is simply the sum of their focal lengths:

$$L = f_o + f_e$$

### The Mathematics of Magnification

The primary purpose of a telescope is to increase the visual angle of the object, making it appear larger. The angular magnification $M$ in normal adjustment is the ratio of the angle subtended by the image to the angle subtended by the object. Mathematically, its magnitude is given by:

$$|M| = \frac{f_o}{f_e}$$

Notice that to achieve a high magnification, the focal length of the objective $f_o$ must be much larger than the focal length of the eyepiece $f_e$.

### Solving the Puzzle

In our specific problem, we are given two crucial pieces of information:
1. The separation between the lenses is $36 	ext{ cm}$, so $f_o + f_e = 36$.
2. The magnitude of angular magnification is $5$, so $\frac{f_o}{f_e} = 5$.

From the magnification equation, we can express the objective's focal length in terms of the eyepiece's focal length:

$$f_o = 5 f_e$$

Now, we substitute this into our length equation:

$$5 f_e + f_e = 36$$

$$6 f_e = 36$$

$$f_e = 6 	ext{ cm}$$

With the focal length of the eyepiece found, we can easily determine the focal length of the objective:

$$f_o = 5 	imes 6 = 30 	ext{ cm}$$

And there we have it! The focal length of the objective is **$30 	ext{ cm}$** and the focal length of the eyepiece is **$6 	ext{ cm}$**. By understanding the physical geometry of normal adjustment, the mathematics naturally falls into place.

Select Answer:

Visualized Solution

$Visualizing Normal Adjustment$

$Length of Telescope$

$Angular Magnification$

$Substituting Values$

$Solving for f_{e}$

$Solving for f_{o}$

$Final Answer$

The Sigma Insight: Optical Instruments

Unlocking the Secrets of the Astronomical Telescope

The Setup

The Mathematics of Magnification

Solving the Puzzle

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