Draw a labelled ray diagram of an astronomical telescope in the near point adjustment position. A giant refracting telescope at an observatory has an objective lens of focal length 15 m and an eyepiece of focal length 1.0 cm. If this telescope is used to view the Moon, find the diameter of the image of the Moon formed by the objective lens. The diameter of the Moon is $3.48\times10^6$ m, and the radius of lunar orbit is $3.8\times10^8$ m.

Draw a labelled ray diagram of an astronomical telescope in the near point adjustment position. A giant refrac

Question

3.48\times10^6

CBSECBSE Class XII Board 2019Subjective· 3mImportance★★★★★

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The telescope’s objective forms a real image of the Moon at its focal plane. Using the small-angle approximation, the image diameter is $d_i = f_o \times \theta$ , where $\theta$ is the Moon’s angular diameter. The result is $d_i \approx 13.7\ \text{cm}$ .

Ray diagram of an astronomical telescope in near point adjustment: the objective forms a real, inverted image A'B' that lies just inside the eyepiece's focal length, so the eyepiece produces a magnified, virtual final image A''B'' at the near point D (25 cm) rather than at infinity.

Concept and intuition

An astronomical telescope works by collecting light from a distant object with a large objective lens (or mirror) and then magnifying the real image formed at the focal plane using an eyepiece. In the near point adjustment (also called normal adjustment for relaxed eye, but here the near point adjustment means the final image is formed at the near point of the eye, 25 cm away), the eyepiece is moved slightly so the final virtual image is at the least distance of distinct vision. However, the question specifically asks for the diameter of the image formed by the objective lens alone — that is the real, intermediate image. This depends only on the objective’s focal length and the angular size of the Moon, not on the eyepiece.

The Moon is so far away that rays from any point on it arrive nearly parallel. The objective lens brings these parallel rays to a focus at its focal plane. The Moon’s angular diameter $\theta$ (the angle it subtends at Earth) is tiny, so the image size is simply $f_o \cdot \theta$ (small-angle approximation: $\tan\theta \approx \theta$ in radians).

Step-by-step solution

1. Find the angular diameter of the Moon

The Moon’s actual diameter $D_m = 3.48 \times 10^6\ \text{m}$ and its distance from Earth (radius of lunar orbit) $R = 3.8 \times 10^8\ \text{m}$ . The angular diameter in radians is:

\theta = \frac{D_m}{R} = \frac{3.48 \times 10^6}{3.8 \times 10^8}

Calculate:

\theta = \frac{3.48}{3.8} \times 10^{-2} \approx 0.9158 \times 10^{-2} = 9.158 \times 10^{-3}\ \text{radians}

Tip

Always use radians for small angles — it directly gives the image size when multiplied by focal length. No need to convert to degrees.

2. Image diameter from the objective

The objective focal length $f_o = 15\ \text{m}$ . The real image of the Moon formed at the focal plane has diameter:

d_i = f_o \cdot \theta = 15 \times 9.158 \times 10^{-3}

d_i = 0.13737\ \text{m} \approx 0.137\ \text{m}

Convert to centimetres:

d_i = 13.7\ \text{cm}

3. Ray diagram for near point adjustment (qualitative)

While the calculation above does not require the diagram, the question asks for it. Here’s how to draw it:

Draw the objective lens (large, convex) on the left. Mark its principal axis. …

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

Showing the 12 most recent of 15 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. An astronomical telescope focussed to infinity has objective lens and eyepiece lens of powers 0.5 diopter and 20 diopter, respectively. Its magnifying power will be$
(a) 8
(b) 20
(c) 30
(d) 40
›Reveal solutionSolution

Focal lengths from the powers give $f_o = 200$ cm, $f_e = 5$ cm; magnifying power $= f_o/f_e = 40$ .

For a lens, power $P$ (in diopters) and focal length $f$ (in metres) are related by $P = 1/f$ .

Objective: $P_o = 0.5$ D $\Rightarrow f_o = 1/0.5 = 2$ m $= 200$ cm.

Eyepiece: $P_e = 20$ D $\Rightarrow f_e = 1/20 = 0.05$ m $= 5$ cm.

…

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CBSE 2025Set 55/6/11 markMCQ
$Q.Assertion (A): In a reflecting telescope, the image does not have chromatic aberration. Reason (R): Chromatic aberration occurs only due to refraction of light through an optical medium. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Both Assertion (A) and Reason (R) are false.$
›Reveal solutionSolution
Chromatic aberration is a refraction-only effect; a reflecting telescope uses mirrors (reflection), so it is free from chromatic aberration. Both statements are true, and the reason correctly explains the assertion. The correct option is (A).

Why This Works — The Concept

Chromatic aberration is the failure of a lens to focus all colours of light to the same point. It happens because the refractive index of glass changes with wavelength — so different colours bend by different amounts when passing through a lens. This is purely a refraction phenomenon.

A reflecting telescope, however, uses a concave mirror as its primary objective. Mirrors work by reflection, not refraction. The law of reflection ( $\angle i = \angle r$ ) is independent of wavelength — all colours reflect at exactly the same angle. So no colour separation occurs, and the image is free from chromatic aberration.

Assertion (A) is true.

In a reflecting telescope (like the Newtonian or Cassegrain design), the primary element is a mirror. Since reflection does not split light into its constituent colours, the image formed has no chromatic aberration. This is a well-known advantage of reflectors over refractors.

Reason (R) is true.

Chromatic aberration arises because the refractive index $n$ of a medium depends on wavelength ( $n(\lambda)$ ). When white light enters a lens, each colour bends by a slightly different angle, causing the focal point to shift with colour. This is a direct consequence of refraction. No refraction — no chromatic aberration.

Is (R) the correct explanation of (A)?

Yes. The reason the reflecting telescope avoids chromatic aberration is precisely because it uses reflection instead of refraction. The reason directly states the physical cause, and that cause is what makes the assertion true. …
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CBSE 2025Set ANNUAL1 mark
$Q. You are given two convex lenses of focal lengths 10 cm and 60 cm. To make a telescope, which of the two lenses will you use as the object lens and which one as the eye lens? OR A convex lens is immersed in water. How is the power of the lens affected?$
›Reveal solutionSolution

The lens with the larger focal length (60 cm) should be the objective and the one with the smaller focal length (10 cm) the eyepiece, since telescope magnification $m = f_o/f_e$ is maximized this way; for the alternative, immersing a convex lens in water reduces (but does not eliminate) its power.

Choosing the objective and eyepiece

For an astronomical (refracting) telescope, the magnifying power (in normal adjustment) is

$m = \frac{f_o}{f_e}$

where $f_o$ is the focal length of the objective and $f_e$ that of the eyepiece. To get the largest possible magnification, $f_o$ should be as large as possible and $f_e$ as small as possible. A large-aperture, large-focal-length objective also gives the telescope better light-gathering power and resolving power.

…

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CBSE 2023Set F1 markMCQ
$Q. The final image formed by a terrestrial telescope is (A) virtual and inverted compared to the object (B) virtual and erect compared to the object (C) real and erect compared to the object (D) none of these$
›Reveal solutionSolution

A terrestrial telescope gives a final image that is virtual and erect relative to the object.

An astronomical telescope forms a final image that is virtual and inverted. A terrestrial telescope is designed to view objects on the ground, where an inverted image is undesirable, so an additional erecting lens is inserted between the objective and the eyepiece. This lens inverts the intermediate image once more, so the final image seen …

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CBSE 2023Set B1 markMCQ
$Q. The focal length of eye piece in telescope ______ the focal length of the objective.$
(i) is less than
(ii) is more than
(iii) is equal
(iv) none of these
›Reveal solutionSolution

For maximum magnification and a compact telescope, the objective has a large focal length and the eyepiece has a short focal length.

The magnifying power of an astronomical telescope (in normal adjustment) is

$m = \frac{f_o}{f_e}$ …

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CBSE 2023Set ANNUAL1 mark
$\dfrac{3\pi}{2}$
›Reveal solutionSolution

For normal adjustment the telescope's magnifying power is the ratio of the objective's to the eyepiece's focal length, giving 5; the alternative converts a phase difference of $3\pi/2$ to a path difference of $3\lambda/4$ .

Solution:

For an astronomical telescope in normal adjustment (final image at infinity), the magnifying power is:

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

Given $f_o = 50\text{ cm}$ , $f_e = 10\text{ cm}$ :

$M = \frac{50}{10} = 5$

Alternative (Or):

Phase difference $\Delta\phi$ and path difference $\Delta x$ are related by: …

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CBSE 2022Set I1 markMCQ
$Q. The magnification power of Astronomical Telescope is (A) f_o/f_e (B) -f_o/f_e (C) -f_e/f_o (D) f_e/f_o$
›Reveal solutionSolution

Magnifying power of astronomical telescope (normal adjustment): M = −f_o/f_e.

An astronomical (refracting) telescope has an objective of focal length f_o and an eyepiece of focal length f_e. In normal adjustment (final image at infinity), the angular magnifying power is:

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

…

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CBSE 2022Set HE2171 mark
$Q. How can the magnification power of Astronomical telescope be increased?$
›Reveal solutionSolution

The magnifying power of an astronomical telescope, $M = f_o/f_e$ , is increased by using a larger objective focal length and/or a smaller eyepiece focal length.

For an astronomical telescope focused for normal (relaxed-eye) viewing, the angular magnifying power is $M = \dfrac{f_o}{f_e}$ , where $f_o$ is the focal length of the objective lens and $f_e$ is the focal length of the eyepiece. To increase $M$ , one can (i) increase the objective's focal length $f_o$ (use a longer-focal-length objective lens/mirror), or (ii) de …

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CBSE 2021Set A1 markMCQ
$Q. The length of an astronomical telescope for normal adjustment is (A) f_o - f_e (B) f_o \times f_e (C) f_o/f_e (D) f_o + f_e$
›Reveal solutionSolution

Length of an astronomical telescope in normal adjustment = f_o + f_e.

In normal adjustment the final image is formed at infinity, so the objective and eyepiece are separated such that the focal point of the objective coincides with the focal point of the eyepiece. The distance between the two lenses (the tube length) is therefore the sum of their focal …

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CBSE 2021Set A1 markMCQ
$Q. Which of the following is correct for Astronomical telescope? (A) f_o = f_e (B) f_o > f_e (C) f_o < f_e (D) f_o << f_e$
›Reveal solutionSolution

For an astronomical telescope f_o > f_e, which gives high magnification.

The angular magnification of an astronomical telescope in normal adjustment is

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

where $f_o$ is the objective focal length and $f_e$ the eyepiece focal length.

…

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CBSE 2020Set ANNUAL1 markMCQ
$Q. If the focal length of objective lens of an astronomical telescope is 20 cm and the length of it is 25cm, the magnification of the telescope for normal adjustment is$
(a) 5
(b) 4
(c) 1.25
(d) 1
›Reveal solutionSolution

In normal adjustment (final image at infinity), an astronomical telescope's tube length equals fo + fe, and its magnifying power is fo/fe.

For an astronomical telescope in normal adjustment, the objective forms an image at its focus which coincides with the focus of the eyepiece, so:

Length of telescope, L = fo + fe

…

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CBSE 2020Set NC1 mark
$Q. A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope?$
›Reveal solutionSolution

The telescope's magnifying power in normal adjustment is simply the ratio of the objective's to the eyepiece's focal length.

Formula

$m = \frac{f_o}{f_e}$

…

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Concept and intuition

Step-by-step solution

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