Larger aperture of objective lens in an astronomical telescope (A) increases the resolving power of telescope.

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Q. Larger aperture of objective lens in an astronomical telescope (A) increases the resolving power of telescope. (B) decreases the brightness of the image. (C) increases the size of the image. (D) decreases the length of the telescope.

CBSECBSE Class XII Board 2020MCQ· 1mImportance★★★★★est

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Concept understanding — Resolving Power of Telescope

Resolving Power of a Telescope – From Intuition to Precision

Imagine you are looking at two stars that are very close together in the night sky. Through a small telescope, they might appear as a single blurry blob. Through a larger telescope, you see them as two distinct points. Why? The answer is not about magnification — it is about resolution, the ability to distinguish fine detail.

The Core Intuition

Light behaves like a wave. When it enters a telescope, it passes through the circular opening (the aperture) and diffracts — it spreads out slightly. Even a perfect lens cannot focus light to an infinitely sharp point. Instead, it produces a small central bright spot surrounded by faint rings, called an Airy pattern. This is not a flaw in the lens; it is a fundamental property of wave optics.

If two stars are extremely close, their Airy patterns overlap. When the overlap is too great, your eye cannot tell that there are two sources — they merge into one. The resolving power of a telescope is its ability to show two closely spaced objects as separate.

The Rayleigh Criterion

There is a precise condition for when two point sources are just barely resolvable. It is called the Rayleigh criterion:

Two sources are resolved when the central maximum of one falls on the first minimum of the other.

For a circular aperture, the angular separation $\Delta\theta$ (in radians) at which this happens is:

\Delta\theta = 1.22 \frac{\lambda}{D}

where:

$\lambda$ is the wavelength of light
$D$ is the diameter of the telescope's objective (the main lens or mirror)

Important

A smaller $\Delta\theta$ means better resolving power — you can distinguish finer details. So a larger $D$ gives a smaller $\Delta\theta$ , which is why big telescopes see more detail.

Why 1.22?

The factor 1.22 comes from the mathematics of diffraction through a circular opening. It is the first zero of the Bessel function that describes the Airy pattern. For a slit (a rectangular opening), the factor would be 1.00 — but telescopes have circular apertures, so 1.22 is the number you need.

A Concrete Example

Suppose you use a telescope with $D = 10$ cm and observe yellow light with $\lambda = 550$ nm ( $5.5 \times 10^{-7}$ m). Then:

\Delta\theta = 1.22 \times \frac{5.5 \times 10^{-7}}{0.1} = 6.71 \times 10^{-6} \text{ radians}

That is about 1.4 arcseconds. Two stars closer than this would appear as one.

Now double the aperture to $D = 20$ cm. The resolving angle halves to $3.35 \times 10^{-6}$ radians (about 0.7 arcseconds). You can now separate stars that are twice as close.

Tip

| Aperture $D$ | Resolving angle $\Delta\theta$ (for $\lambda = 550$ nm) |

|--------------|--------------------------------------------------------|

| 5 cm | 2.8 arcseconds |

| 10 cm | 1.4 arcseconds |

| 20 cm | 0.7 arcseconds |

| 1 m | 0.14 arcseconds |

What Resolving Power Is NOT …

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A larger objective lens gathers more light and reduces the diffraction limit, directly increasing resolving power while also brightening the image. The answer is (A).

Why the objective lens aperture matters

The objective lens is the first optical element light encounters in a telescope. Its diameter—the aperture—controls two fundamental aspects of image formation: how much light the telescope collects (affecting brightness) and how finely it can distinguish nearby point sources (resolving power). Understanding these dependencies reveals why aperture is the single most important specification in telescope design.

Resolving power measures the telescope's ability to separate two closely spaced objects. According to the Rayleigh criterion, two point sources are just resolved when the central maximum of one diffraction pattern falls on the first minimum of the other. For a circular aperture, this angular separation is:

\theta_{\text{min}} = 1.22 \frac{\lambda}{D}

where $D$ is the objective diameter and $\lambda$ is the wavelength of light. Resolving power is defined as the reciprocal of this minimum angle:

R = \frac{1}{\theta_{\text{min}}} = \frac{D}{1.22\lambda}

So resolving power is directly proportional to aperture. A larger $D$ means smaller $\theta_{\text{min}}$ , allowing the telescope to distinguish finer details.

Examining each option

Option (A): Increases resolving power

From the formula above, doubling the aperture doubles the resolving power. A 200 mm objective can separate stars twice as close together as a 100 mm objective can. This is the primary reason astronomers prize large telescopes.
Option (B): Decreases brightness

Light-gathering power is proportional to the area of the objective, hence to $D^2$ . A larger aperture collects more photons per unit time, making the image brighter, not dimmer. This option contradicts basic optics.
Option (C): Increases the size of the image …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2022Set M1 mark
$Q. How does the resolving power of a telescope change on increasing the diameter of the objective lens?$
›Reveal solutionSolution

Resolving power ∝ diameter of objective, so it increases.

The resolving power of a telescope is the ability to distinguish two close objects as separate. It is given by
$\text{R.P.} = \frac{D}{1.22\,\lambda}$
where $D$ is the diameter (aperture) of the objective lens and $\lambda$ is the wavelength of light. …
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CBSE 2020Set 55/1/11 markMCQ
$Q. Larger aperture of objective lens in an astronomical telescope (A) increases the resolving power of telescope. (B) decreases the brightness of the image. (C) increases the size of the image. (D) decreases the length of the telescope.$
›Reveal solutionSolution
A larger objective lens gathers more light and reduces the diffraction limit, directly increasing resolving power while also brightening the image. The answer is (A).

Why the objective lens aperture matters

The objective lens is the first optical element light encounters in a telescope. Its diameter—the aperture—controls two fundamental aspects of image formation: how much light the telescope collects (affecting brightness) and how finely it can distinguish nearby point sources (resolving power). Understanding these dependencies reveals why aperture is the single most important specification in telescope design.

Resolving power measures the telescope's ability to separate two closely spaced objects. According to the Rayleigh criterion, two point sources are just resolved when the central maximum of one diffraction pattern falls on the first minimum of the other. For a circular aperture, this angular separation is:
$\theta_{\text{min}} = 1.22 \frac{\lambda}{D}$
where $D$ is the objective diameter and $\lambda$ is the wavelength of light. Resolving power is defined as the reciprocal of this minimum angle:
$R = \frac{1}{\theta_{\text{min}}} = \frac{D}{1.22\lambda}$
So resolving power is directly proportional to aperture. A larger $D$ means smaller $\theta_{\text{min}}$ , allowing the telescope to distinguish finer details.

Examining each option

Option (A): Increases resolving power

From the formula above, doubling the aperture doubles the resolving power. A 200 mm objective can separate stars twice as close together as a 100 mm objective can. This is the primary reason astronomers prize large telescopes.

Option (B): Decreases brightness

Light-gathering power is proportional to the area of the objective, hence to $D^2$ . A larger aperture collects more photons per unit time, making the image brighter, not dimmer. This option contradicts basic optics.

Option (C): Increases the size of the image …
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CBSE 2020Set ANNUAL1 mark
$0.5\,\mu m$
›Reveal solutionSolution

$\Delta\theta = 1.22\lambda/D \approx 2.5\times10^{-7}\,rad$ , using aperture diameter $D = 2\times(\text{radius})$ .

The limit of angular resolution (minimum resolvable angle) of a telescope, set by diffraction at its circular objective, is

$\Delta\theta = \frac{1.22\,\lambda}{D}$

where $D$ is the diameter of the objective aperture and $\lambda$ is the wavelength of light.

Given $\lambda = 0.5\,\mu m = 5\times10^{-7}\,m$ and the objective's radius $= 122\,cm = 1.22\,m$ , the aperture diameter is

$D = 2 \times 1.22\,m = 2.44\,m$

$\Delta\theta = \frac{1.22 \times 5\times10^{-7}}{2.44} = \frac{6.1\times10^{-7}}{2.44} = 2.5\times10^{-7}\,rad$

…

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CBSE 2018Set ANNUAL1 markMCQ
$Q. The resolving power of human eye (in minute) is -$
(a) 1/60
(b) 1
(c) 10
(d) 1/2
›Reveal solutionSolution

The eye's limit of resolution is roughly 1 arc-minute.

The smallest angular separation the normal human eye can distinguish is set by diffraction at the pupil and the spacing of retinal cells, and comes out to about one minute of arc (1′ ≈ 1/60 of a degree ≈ …

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Q.Larger aperture of objective lens in an astronomical telescope (A) increases the resolving power of telescope. (B) decreases the brightness of the image. (C) increases the size of the image. (D) decreases the length of the telescope.

Why the objective lens aperture matters

Examining each option

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