An astronomical telescope has a large aperture to
(a) reduce spherical aberration
(b) have high reso

Q: Why can't we just use a higher magnification eyepiece to see better details?

Magnification only enlarges the image. If the image is already blurred due to diffraction (overlapping Airy disks), magnifying it will just give you a larger blurred image. Resolution is what actually provides the fine detail, and that depends strictly on the aperture.

Q: What exactly is the 'aperture' of a telescope?

The aperture is the clear diameter of the main light-gathering element of the telescope. For a refracting telescope, it's the diameter of the objective lens. For a reflecting telescope, it's the diameter of the primary mirror.

Q: Does a larger aperture help with anything else besides resolution?

Yes! A larger aperture also has a larger surface area, which means it can gather much more light. This allows astronomers to see much fainter and more distant objects in the universe.

Q: Where does the number 1.22 come from in the formula?

The factor 1.22 comes from the mathematical calculation of the diffraction pattern of a circular aperture using Bessel functions. It represents the position of the first dark ring (minimum) surrounding the central bright Airy disk.

Q: Does a large aperture reduce spherical aberration?

No, actually, larger lenses are more prone to spherical aberration because the outer edges of a large spherical lens bend light more strongly than the center. Special parabolic shapes or corrective lenses are needed to fix this.

Question

An astronomical telescope has a large aperture to

Options:

(A) reduce spherical aberration

(B) have high resolution

(C) increase span of observation

(D) have low dispersion

Sigma Percentile · Accepted Answer

### The Quest for the Stars

Since the dawn of humanity, we have looked up at the night sky, wondering about the celestial bodies scattered across the vast cosmic ocean. To get a closer look, we invented the astronomical telescope. When you look at a distant binary star system (two stars orbiting each other), they often appear as a single, blurry point of light to the naked eye. The primary job of a telescope is not just to make things look bigger, but to make them look *clearer*. This ability to distinguish two closely spaced objects as separate entities is known as **Resolving Power**.

### The Enemy: Diffraction

You might think that if we just use a more powerful eyepiece to increase the magnification, we could see the two stars clearly. However, physics throws a curveball at us in the form of **diffraction**.

When light from a distant star enters the circular aperture (the objective lens or mirror) of a telescope, it doesn't just travel in straight lines. The light waves bend around the edges of the aperture, creating a diffraction pattern. Instead of a perfect point of light, the star is imaged as a central bright disk surrounded by faint concentric rings. This is called the **Airy disk**.

If two stars are very close to each other, their Airy disks will overlap. If they overlap too much, they blend into a single blob, and no amount of magnification will separate them. Magnifying a blurry blob just gives you a bigger blurry blob!

### Rayleigh's Criterion and Resolving Power

Lord Rayleigh proposed a criterion for when two objects are "just resolved." According to **Rayleigh's Criterion**, two point sources are just resolved when the central maximum of one diffraction pattern coincides with the first minimum of the other.

Mathematically, the angular separation $\Delta 	heta$ between the two stars must be at least:

$$ \Delta 	heta = \frac{1.22 \lambda}{d} $$

Where:
- $\lambda$ is the wavelength of the light.
- $d$ is the diameter of the objective lens (the aperture).

The **Resolving Power (RP)** of a telescope is defined as the reciprocal of this minimum angular separation. It tells us how well the telescope can resolve fine details.

$$ RP = \frac{1}{\Delta 	heta} = \frac{d}{1.22 \lambda} $$

### The Master Equation: Why Bigger is Better

Let's look closely at our master equation:

$$ RP = \frac{d}{1.22 \lambda} $$

This elegant formula reveals a profound truth about optical instruments. The resolving power is **directly proportional** to the aperture $d$.

$$ RP \propto d $$

If you want to see finer details—whether it's the rings of Saturn, the Great Red Spot on Jupiter, or separating a tight binary star system—you need a higher resolving power. And the only way to achieve that (assuming you are observing in visible light, so $\lambda$ is fixed) is to increase the diameter $d$ of your objective lens or mirror.

### The Verdict

This is the exact reason why astronomical observatories are built with massive mirrors, some spanning over 10 meters across! A large aperture does two critical things: it gathers more light (making faint objects visible) and it drastically increases the resolving power (making blurry objects sharp).

Therefore, an astronomical telescope has a large aperture primarily to **have high resolution**.

*(Note: While a large aperture gathers more light, it does not inherently reduce spherical aberration or dispersion; in fact, larger lenses are harder to manufacture without aberrations. The primary fundamental advantage dictated by wave optics is the increase in resolving power.)*

Animated Solution for Physics - Optics: An astronomical telescope has a large aperture to

Select Answer:

Visualized Solution

Objective Lens of a Telescope

Resolving Power Formula

Conclusion

The Sigma Insight: Optical Instruments

The Quest for the Stars

The Enemy

Rayleigh's Criterion and Resolving Power

The Master Equation

The Verdict

Similar Questions

Some laws/processes are given in Column I. Match these with the physical phenomena given in Column II.

List-I

List-II

The aperture of a telescope is $5 m$ . The separation between the moon and the earth is $4 \times 1 0^{5} km$ . With light of wavelength of $5500 \overset{˚}{A}$ , the minimum separation between objects on the surface of moon, so that they are just resolved, is close to

Diameter of the objective lens of a telescope is $250 cm$ . For light of wavelength $600 nm$ coming from a distant object, the limit of resolution of the telescope is close to

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

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

A planet is observed by an astronomical refracting telescope having an objective of focal length $16 m$ and an eyepiece of focal length $2 cm$

Calculate the limit of resolution of a telescope objective having a diameter of $200 cm$ , if it has to detect light of wavelength $500 nm$ coming from a star.

The magnifying power of a telescope with tube length $60 cm$ is $5$ . What is the focal length of its eyepiece?

An observer looks at a distance tree of height 10 m with a telescope of magnifying power of 20. To the observer the tree appears

The image formed by an objective of a compound microscope is