(a) For the telescope described in Exercise 9.27 (a), what is the separation between the objective lens and th

Exercises · 9.28

Q. (a) For the telescope described in Exercise 9.27 (a), what is the separation between the objective lens and the eyepiece?

(b) If this telescope is used to view a

100\ \text{m}

tall tower

3\ \text{km}

away, what is the height of the image of the tower formed by the objective lens?

25\ \text{cm}

CBSENCERTSubjective· 3mImportance★★★★★

49% · 36/73 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

For a telescope in normal adjustment, the separation between objective and eyepiece equals the sum of their focal lengths. The objective forms a real, inverted image of a distant object at its focal plane; using the angular magnification and the near-point distance gives the final image height.

Separation between objective and eyepiece For a telescope used in normal adjustment (the final image is at infinity), the eyepiece is placed so that its first focal plane coincides with the focal plane of the objective. This means the distance between the two lenses is simply: $\text{Separation} = f_o + f_e$ From Exercise 9.27 (a), we have $f_o = 140\ \text{cm}$ and $f_e = 5.0\ \text{cm}$ . Therefore: $\text{Separation} = 140 + 5.0 = 145\ \text{cm}$
Watch out

A common mistake is to use the formula for a microscope (separation = $f_o + f_e$ only when the final image is at infinity). For a telescope, this is correct in normal adjustment — but if the final image is formed at the near point, the separation changes slightly. Here, part (a) assumes normal adjustment.
Height of the image formed by the objective lens The tower is $100\ \text{m}$ tall and located $3\ \text{km} = 3000\ \text{m}$ away. For a distant object, the objective lens forms a real, inverted image at its focal plane. The angular size of the object as seen from the objective is: $\theta = \frac{\text{height of object}}{\text{distance}} = \frac{100}{3000} = \frac{1}{30}\ \text{radian}$ Since the image lies at the focal plane, its height $h_o$ is given by: $h_o = f_o \cdot \theta = 140\ \text{cm} \times \frac{1}{30} = \frac{140}{30}\ \text{cm} \approx 4.67\ \text{cm}$
Tip

The relation $h = f \theta$ works because for small angles, $\tan \theta \approx \theta$ , and the image distance is essentially $f_o$ for a distant object. This is the same principle used in camera lenses.

So the objective forms an image about $4.67\ \text{cm}$ tall (inverted). (c) Height of the final image when formed at $25\ \text{cm}$ The eyepiece acts as a simple magnifier. The intermediate image (from the objective) is placed just inside the focal point of the eyepiece so that the final image is virtual and at the near point ( $25\ \text{cm}$ from the eye). First, find the magnification produced by the eyepiece. For a lens used as a magnifier with the final image at the near point: $m_e = 1 + \frac{D}{f_e}$ where $D = 25\ \text{cm}$ is the least distance of distinct vision. So:

Start your 14-day free trial to unlock the full solution →

Method: Two-Step Image Formation in a Telescope (Objective → Eyepiece)

This method treats the telescope as two independent lenses in sequence. The objective forms a real, inverted image at its focal plane; the eyepiece then magnifies that image.

(a) Separation between objective and eyepiece

Step 1 – Recall the normal adjustment condition

For a telescope in normal adjustment, the final image is at infinity. This requires that the image from the objective lies exactly at the focal point of the eyepiece.

Step 2 – Write the separation formula

Separation $L$ = distance between the two lenses = $f_o + f_e$

Step 3 – Use data from Exercise 9.27(a)

From that exercise (standard NCERT values):

Focal length of objective, $f_o = 140\ \text{cm}$
Focal length of eyepiece, $f_e = 5\ \text{cm}$

Step 4 – Calculate

L = 140 + 5 = 145\ \text{cm}

Answer: $\boxed{145\ \text{cm}}$

(b) Height of image formed by objective lens

Step 1 – Identify the relevant formula

The objective acts like a simple lens. For a distant object, the image forms at the focal plane. Use the lens formula or magnification formula:

\text{Magnification of objective} = \frac{h_i}{h_o} = \frac{v}{u}

For a very distant object, $u \approx \infty$ , so $v \approx f_o$ .

Step 2 – Use the small-angle approximation

Alternatively, use the direct relation:

\frac{h_i}{f_o} = \frac{h_o}{u}

where $h_o$ = object height, $u$ = object distance.

Step 3 – Substitute values

$h_o = 100\ \text{m}$
$u = 3\ \text{km} = 3000\ \text{m}$
$f_o = 140\ \text{cm} = 1.4\ \text{m}$

h_i = h_o \times \frac{f_o}{u} = 100 \times \frac{1.4}{3000}

Step 4 – Compute

h_i = \frac{140}{3000} = 0.04667\ \text{m} = 4.67\ \text{cm}

Answer: $\boxed{4.67\ \text{cm}}$

(c) Height of final image (formed at $25\ \text{cm}$ )

Step 1 – Treat the eyepiece as a simple magnifier

The image from the objective ( $h_i = 4.67\ \text{cm}$ ) acts as the object for the eyepiece. This object is placed at the focal plane of the objective, which is at distance $f_e$ from the eyepiece. …

Start your 14-day free trial to unlock the full solution →

Here are the common mistakes students make on this microscope/telescope magnification problem, along with how to avoid each.

(a) Separation between objective and eyepiece

✗ Mistake 1: Using the wrong formula for separation

Students often confuse the microscope formula ( $L = v_o + f_e$ ) with the telescope formula.

How to avoid:

For a telescope in normal adjustment, the separation is simply:

L = f_o + f_e

The image from the objective is formed at its focal point, and the eyepiece is placed so its first focal point coincides with that image.

Key: No $v_o$ or $u_o$ — just add the two focal lengths.

✗ Mistake 2: Forgetting to convert units

Focal lengths may be given in cm, but the answer is expected in metres or cm consistently.

How to avoid:

Always write the given values with units, convert to a single unit before adding, and state the final answer with the correct unit.

(b) Height of image formed by the objective lens

✗ Mistake 3: Using the telescope magnification formula directly

Some students try to use $M = f_o / f_e$ to find the image height — that gives the angular magnification, not the linear height from the objective.

How to avoid:

The objective lens acts like a simple camera. Use the lens formula and magnification formula:

Object distance $u = -3000\ \text{m}$ (negative by sign convention)
Focal length $f_o$ (from part (a) data)
Find $v$ using:

\frac{1}{v} = \frac{1}{f_o} + \frac{1}{u}

Then linear magnification:

m_o = \frac{v}{u}

Image height:

h_i = m_o \times h_o

Key: The tower is very far ( $u \gg f_o$ ), so $v \approx f_o$ . This approximation is valid but show the exact step first.

✗ Mistake 4: Sign convention errors

Using $u = +3000$ or forgetting the negative sign leads to a wrong $v$ .

How to avoid:

For a convex lens, real object distance is negative in the Cartesian sign convention. Always write:

u = -3\ \text{km} = -3000\ \text{m}

(c) Height of the final image at $25\ \text{cm}$

✗ Mistake 5: Confusing near point with image distance

Students often set $v_e = -25\ \text{cm}$ (correct) but then forget the sign when using the lens formula.

How to avoid:

The final image is virtual and on the same side as the object for the eyepiece. So:

v_e = -25\ \text{cm}

Use the lens formula for the eyepiece:

\frac{1}{u_e} = \frac{1}{v_e} - \frac{1}{f_e}

Then find $m_e = v_e / u_e$ .

✗ Mistake 6: Multiplying objective and eyepiece magnifications incorrectly

Some students multiply $m_o$ (from part b) by $m_e$ but forget that $m_o$ is already a linear magnification.

How to avoid:

Total linear magnification:

m_{\text{total}} = m_o \times m_e

Then:

$h_{\text{final}} = m_{\text{total}} \times h_o$ …

Start your 14-day free trial to unlock the full solution →

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.

…

Start your 14-day free trial to see the full solution →
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. …
Start your 14-day free trial to see the full solution →
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.

…

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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

$m = \frac{f_o}{f_e}$ …

Start your 14-day free trial to see the full solution →
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: …

Start your 14-day free trial to see the full solution →
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:

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

…

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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.

…

Start your 14-day free trial to see the full solution →
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

…

Start your 14-day free trial to see the full solution →
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

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

…

Start your 14-day free trial to see the full solution →

←Previous question9.27 — A small telescope has an objective lens of focal length 140cm and an e…→Next question9.29 — A Cassegrain telescope uses two mirrors as shown in Fig. 9.26. Such a …

Q.(a) For the telescope described in Exercise 9.27 (a), what is the separation between the objective lens and the eyepiece?

Concept understanding — Astronomical Telescope

Method: Two-Step Image Formation in a Telescope (Objective → Eyepiece)

(a) Separation between objective and eyepiece

(b) Height of image formed by objective lens

(c) Height of final image (formed at $25\ \text{cm}$ )

(a) Separation between objective and eyepiece

✗ Mistake 1: Using the wrong formula for separation

✗ Mistake 2: Forgetting to convert units

(b) Height of image formed by the objective lens

✗ Mistake 3: Using the telescope magnification formula directly

✗ Mistake 4: Sign convention errors

(c) Height of the final image at $25\ \text{cm}$

✗ Mistake 5: Confusing near point with image distance

✗ Mistake 6: Multiplying objective and eyepiece magnifications incorrectly

Why This Works — The Concept

Q.(a) For the telescope described in Exercise 9.27 (a), what is the separation between the objective lens and the eyepiece?

🎓Unlock everything free for 14 days

Unlock everything free for 14 days