A small telescope has an objective lens of focal length $140\ \text{cm}$ and an eyepiece of focal length $5.0\ \text{cm}$. What is the magnifying power of the telescope for viewing distant objects when (a) the telescope is in normal adjustment (i.e., when the final image is at infinity)? (b) the final image is formed at the least distance of distinct vision ($25\ \text{cm}$)?

A small telescope has an objective lens of focal length 140cm and an eyepiece of focal length 5.0cm. What is t

Exercises · 9.27

140\ \text{cm}

(a) the telescope is in normal adjustment (i.e., when the final image is at infinity)?

(b) the final image is formed at the least distance of distinct vision (

25\ \text{cm}

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The magnifying power of a telescope is the ratio of the angle subtended by the image to the angle subtended by the object. For normal adjustment, it equals $f_o / f_e = 28$ ; for the final image at the near point, it is $f_o / f_e (1 + f_e / D) = 33.6$ .

Why Magnification Works This Way

A telescope makes distant objects appear larger by increasing the angular size of the image compared to the object. The objective lens forms a real, inverted image of the distant object at its focal plane. The eyepiece then acts like a magnifying glass to view that intermediate image. The magnifying power $M$ is defined as:

M = \frac{\text{angle subtended by the final image at the eye}}{\text{angle subtended by the object at the unaided eye}}

For a distant object, the angle subtended at the unaided eye is essentially the same as the angle subtended at the objective. The trick is that the eyepiece lets you bring the intermediate image much closer to your eye, making it appear under a larger angle.

M_\infty = \frac{f_o}{f_e}

Let’s apply these directly.

Step-by-Step Solution

Given:

$f_o = 140\ \text{cm}$ , $f_e = 5.0\ \text{cm}$ , least distance of distinct vision $D = 25\ \text{cm}$ .

1. Normal adjustment (final image at infinity)

In normal adjustment, the eyepiece is adjusted so that the intermediate image lies exactly at its focal point. The eyepiece then produces parallel rays (image at infinity), which the relaxed eye views without strain. The magnifying power is simply the ratio of focal lengths:

M_\infty = \frac{f_o}{f_e} = \frac{140}{5.0} = 28

So the telescope magnifies 28 times.

Tip

Normal adjustment gives the least magnification for a given telescope, but it is the most comfortable for the eye because the ciliary muscles are relaxed.

2. Final image at the least distance of distinct vision ( $D = 25\ \text{cm}$ ) …

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Method: Magnification Formula for a Refracting Telescope

We use the standard magnification equations for a telescope in two common viewing configurations.

Key Concept

For a telescope, the magnifying power $M$ is the ratio of the angle subtended by the final image at the eye to the angle subtended by the object at the unaided eye. For distant objects, this simplifies to:

Normal adjustment (final image at infinity):

M = -\frac{f_o}{f_e}

where $f_o$ = focal length of objective, $f_e$ = focal length of eyepiece.

The negative sign indicates an inverted image (not always required in magnitude).

Final image at near point (least distance of distinct vision $D = 25\ \text{cm}$ ):

M = -\frac{f_o}{f_e} \left(1 + \frac{f_e}{D}\right)

Step-by-Step Solution

Given:

$f_o = 140\ \text{cm}$ , $f_e = 5.0\ \text{cm}$ , $D = 25\ \text{cm}$

(a) Normal adjustment (image at infinity)

Step 1: Apply the formula

M = -\frac{f_o}{f_e}

Step 2: Substitute values

M = -\frac{140}{5.0} = -28

Step 3: Interpret

The magnifying power is 28 (magnitude). The negative sign means the image is inverted relative to the object.

Answer (a): $\boxed{28}$ (or $|M| = 28$ )

(b) Final image at least distance of distinct vision ( $25\ \text{cm}$ )

Step 1: Apply the formula for near-point viewing

M = -\frac{f_o}{f_e} \left(1 + \frac{f_e}{D}\right)

Step 2: Substitute values

M = -\frac{140}{5.0} \left(1 + \frac{5.0}{25}\right)

Step 3: Simplify inside the bracket

1 + \frac{5.0}{25} = 1 + 0.2 = 1.2

Step 4: Multiply

$M = -28 \times 1.2 = -33.6$ …

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Here are the common mistakes students make on this telescope magnification problem, and how to avoid each.

1. Using the wrong formula for each case

The Mistake:

Students often use the same formula for both parts. For normal adjustment (final image at infinity), the magnifying power is:

M = \frac{f_o}{f_e}

For the final image at the near point ( $D = 25\ \text{cm}$ ), the correct formula is:

M = \frac{f_o}{f_e} \left(1 + \frac{f_e}{D}\right)

Some students incorrectly use $M = \frac{f_o}{f_e} + \frac{D}{f_e}$ or forget the $+1$ term.

How to Avoid:

Memorise the two distinct cases:
- Normal adjustment → $M = \frac{f_o}{f_e}$ (image at infinity, relaxed eye)
- Near point adjustment → $M = \frac{f_o}{f_e} \left(1 + \frac{f_e}{D}\right)$ (image at $D$ , maximum strain)
Write the formula explicitly before substituting numbers.

2. Confusing objective and eyepiece focal lengths

The Mistake:

Swapping $f_o$ and $f_e$ in the formula. For a telescope, $f_o$ is large (objective) and $f_e$ is small (eyepiece). Using $f_o = 5\ \text{cm}$ and $f_e = 140\ \text{cm}$ gives a magnifying power less than 1 — physically impossible for a telescope.

How to Avoid:

Always label: $f_o = 140\ \text{cm}$ (objective), $f_e = 5.0\ \text{cm}$ (eyepiece).
Check: Magnification > 1 for a telescope. If your answer is < 1, you’ve swapped them.

3. Forgetting to convert units

The Mistake:

Using $f_o = 140\ \text{cm}$ and $f_e = 5.0\ \text{cm}$ directly in the formula — that’s fine here because both are in cm. But if $D$ is given in cm and $f_e$ in metres, or vice versa, students forget to convert.

How to Avoid:

Ensure all lengths are in the same unit before substituting. Here, $D = 25\ \text{cm}$ matches $f_e = 5.0\ \text{cm}$ , so no conversion needed.
If units differ, convert everything to cm or metres consistently.

4. Misinterpreting “normal adjustment”

The Mistake:

Thinking “normal adjustment” means the final image is at the near point ( $25\ \text{cm}$ ). Actually, normal adjustment means the final image is at infinity — the eye is relaxed.

How to Avoid:

Memorise the definitions:
- Normal adjustment → final image at infinity → $M = f_o / f_e$
- Near point adjustment → final image at $D$ → $M = \frac{f_o}{f_e} \left(1 + \frac{f_e}{D}\right)$
Read the question carefully: “when the telescope is in normal adjustment (i.e., when the final image is at infinity)” — the hint is already given.

5. Sign errors in the formula derivation

The Mistake:

Some textbooks derive magnification with a negative sign (indicating inverted image). Students sometimes include the negative sign in the final answer, or forget it when the question asks for “magnifying power” (which is usually taken as positive).

How to Avoid:

Magnifying power is generally reported as a positive number (absolute value). The negative sign indicates inversion, but exam questions typically want the magnitude.
If the question asks for “magnifying power,” give $|M|$ .

6. Arithmetic errors in the second part

The Mistake: …

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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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Q.A small telescope has an objective lens of focal length $140\ \text{cm}$ and an eyepiece of focal length $5.0\ \text{cm}$ . What is the magnifying power of the telescope for viewing distant objects when

Concept understanding — Astronomical Telescope

Why Magnification Works This Way

Step-by-Step Solution

1. Normal adjustment (final image at infinity)

2. Final image at the least distance of distinct vision ( $D = 25\ \text{cm}$ ) …

Method: Magnification Formula for a Refracting Telescope

Key Concept

Step-by-Step Solution

(a) Normal adjustment (image at infinity)

(b) Final image at least distance of distinct vision ( $25\ \text{cm}$ )

1. Using the wrong formula for each case

2. Confusing objective and eyepiece focal lengths

3. Forgetting to convert units

4. Misinterpreting “normal adjustment”

5. Sign errors in the formula derivation

6. Arithmetic errors in the second part

Why This Works — The Concept

Q.A small telescope has an objective lens of focal length 140 cm140\ \text{cm}140 cm and an eyepiece of focal length 5.0 cm5.0\ \text{cm}5.0 cm. What is the magnifying power of the telescope for viewing distant objects when

Why Magnification Works This Way

Step-by-Step Solution

1. Normal adjustment (final image at infinity)

2. Final image at the least distance of distinct vision (D=25 cmD = 25\ \text{cm}D=25 cm) …

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Q.A small telescope has an objective lens of focal length $140\ \text{cm}$ and an eyepiece of focal length $5.0\ \text{cm}$ . What is the magnifying power of the telescope for viewing distant objects when

2. Final image at the least distance of distinct vision ( $D = 25\ \text{cm}$ ) …

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