Compound microscope: M = (L/f_o)(D/f_e), where L is tube length, D = 25 cm (least distinct vision). Astronomical telescope (normal adj): M = -f_o/f_e.
-- NCERT Class 12 Physics, Ch. 9, p. 244Astronomical Telescope
8 MCQs1 revision card9-step worked example
Source: NCERT OpticsPYQ coverage: NEET 2020, 2024Official key: NTA-verifiedLast updated: 27 Sep 2026
Astronomical Telescope, explained for NEET
The telescope formula has exactly one high-frequency error mode, and the corpus names it: inverting f_o and f_e. Writing M = f_e/f_o instead of f_o/f_e turns a magnification of 20 into 0.05 — and papers reliably place both values among the options.
An astronomical telescope pairs a long-focal-length objective with a short-focal-length eyepiece. The objective collects light from an object at infinity and forms a real, inverted image at its focal plane. The eyepiece then acts as a simple magnifier on that image. In normal adjustment, the intermediate image sits at the common focus of both lenses, so the final image is formed at infinity and the eye views it relaxed.
NCERT Class 12 Physics Part 2, Chapter 9, page 244 gives the two results you need together:
- Magnifying power: M = −f_o/f_e
- Tube length: L = f_o + f_e
The minus sign encodes the inverted final image. For astronomical objects that inversion doesn't matter, which is why no erecting lens is used — that is the design difference from a terrestrial telescope.
Read the ratio physically before you compute. Large magnification needs f_o large and f_e small. If your answer says a 100 cm objective with a 5.0 cm eyepiece gives M = 0.05, the ratio is upside down. This single sanity check kills the dominant distractor.
Reflecting vs refracting: a refracting telescope uses a converging lens as objective; a reflecting telescope uses a concave mirror. The mirror version avoids chromatic aberration entirely (mirrors don't disperse), can be built with much larger apertures, and gathers more light — the reason all large research telescopes are reflectors. The magnification relation M = −f_o/f_e is unchanged; f_o is simply the mirror's focal length.
Watch out: normal adjustment is an assumption, not a default. When a question places the final image at the near point instead, L = f_o + f_e no longer holds.
Can you answer these Astronomical Telescope MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In an astronomical telescope in normal adjustment, the final image is formed at
Show answer and why every option is right or wrong
Answer: B. B is correct. Normal adjustment is defined by the final image being at infinity, so the eye views it without accommodation — NCERT Class 12 Physics Part 2, Chapter 9, page 244.
Why A is wrong: A is wrong because the objective's focal point is where the intermediate real image forms, not the final image the eye sees.
Why C is wrong: C is wrong because near-point viewing is the alternative adjustment, not the normal one; it changes both the magnification expression and the tube length.
Why D is wrong: D is wrong because no image is formed at a lens's optical centre; the optical centre is a reference point for ray construction.
MCQ 2Direct ApplicationPractice
A refracting astronomical telescope has an objective of focal length 1.20 m and an eyepiece of focal length 2.00 × 10⁻² m. Its magnifying power in normal adjustment has magnitude
Show answer and why every option is right or wrong
Answer: A. A is correct. |M| = f_o/f_e = 1.20 / (2.00 × 10⁻²) = 60.0, applying M = −f_o/f_e from NCERT Class 12 Physics Part 2, Chapter 9, page 244.
Why B is wrong: B is wrong because it multiplies the two focal lengths (1.20 × 2.00 × 10⁻² scaled), which has no basis in the magnification relation.
Why C is wrong: C is wrong because it is f_e/f_o — the inverted ratio. A telescope built to magnify cannot have |M| far below 1; this is the pattern's standard distractor.
Why D is wrong: D is wrong because it adds the focal lengths (1.20 + 2.00 × 10⁻² = 1.22 m). That sum is the tube length in metres, not the magnifying power.
MCQ 3Easy RecallPractice
The objective of a reflecting astronomical telescope is
Show answer and why every option is right or wrong
Answer: D. D is correct. A reflecting telescope uses a concave mirror as its light-collecting objective; a refracting telescope uses a converging lens instead — NCERT Class 12 Physics Part 2, Chapter 9, page 244.
Why A is wrong: A is wrong because a concave lens diverges light and cannot form the real image the eyepiece needs.
Why B is wrong: B is wrong because a convex lens objective defines the refracting telescope, which is the other design.
Why C is wrong: C is wrong because a plane mirror has no focal length and cannot converge light to form the intermediate image.
MCQ 4Direct ApplicationPractice
An astronomical telescope in normal adjustment has a tube length of 1.05 m, and its eyepiece has focal length 5.0 × 10⁻² m. The focal length of the objective is
Show answer and why every option is right or wrong
Answer: C. C is correct. In normal adjustment L = f_o + f_e, so f_o = 1.05 − 5.0 × 10⁻² = 1.00 m — NCERT Class 12 Physics Part 2, Chapter 9, page 244.
Why A is wrong: A is wrong because it adds the eyepiece focal length instead of subtracting it; the tube already contains f_e.
Why B is wrong: B is wrong because 0.21 m is 1.05/5.0: it divides the tube length by the eyepiece focal length taken in centimetres instead of subtracting it. In normal adjustment L = f_o + f_e, so f_o = 1.05 − 0.05 = 1.00 m.
Why D is wrong: D is wrong because it multiplies L by f_e, which produces a quantity with units of m² and cannot be a focal length.
MCQ 5Concept TrapPractice
Two telescope designs use the same eyepiece. Design X has an objective of focal length 2.00 m; design Y has an objective of focal length 0.50 m. Compared with Y, design X has
Show answer and why every option is right or wrong
Answer: D. D is correct. |M| = f_o/f_e rises with f_o, and L = f_o + f_e rises with f_o too — so the higher-power astronomical telescope is also the longer one — NCERT Class 12 Physics Part 2, Chapter 9, page 244.
Why A is wrong: A is wrong on both counts: a larger f_o raises the ratio f_o/f_e and lengthens the tube.
Why B is wrong: B is wrong because it inverts the magnification dependence — it treats f_o as if it sat in the denominator.
Why C is wrong: C is wrong because it correctly raises the magnification but wrongly shortens the tube; L = f_o + f_e grows with f_o.
MCQ 6Easy RecallPractice
A principal advantage of a reflecting telescope over a refracting telescope of the same aperture is that the reflector
Show answer and why every option is right or wrong
Answer: A. A is correct. A mirror reflects all wavelengths through the same geometry, so it does not disperse light and the objective contributes no chromatic aberration — NCERT Class 12 Physics Part 2, Chapter 9, page 244.
Why B is wrong: B is wrong because both designs give an inverted final image for astronomical use; neither erects it, which is exactly why M carries a minus sign.
Why C is wrong: C is wrong because M = −f_o/f_e depends on the eyepiece in both designs; only the nature of the objective changes.
Why D is wrong: D is wrong because both designs assume a distant object; the objective forms the intermediate image at its focal plane in either case.
MCQ 7CalculationPractice
An astronomical telescope in normal adjustment has a magnifying power of magnitude 15.0 and a tube length of 0.80 m. The focal length of its eyepiece is
Show answer and why every option is right or wrong
Answer: B. B is correct. From |M| = f_o/f_e, f_o = 15.0 f_e; substituting into L = f_o + f_e gives 0.80 = 16.0 f_e, so f_e = 5.0 × 10⁻² m — NCERT Class 12 Physics Part 2, Chapter 9, page 244.
Why A is wrong: A is wrong because it divides L by 15.0 instead of by 16.0, forgetting that the tube length contains f_e as well as f_o.
Why C is wrong: C is wrong because 0.75 m is f_o (= 15.0 × 5.0 × 10⁻²); the question asks for the eyepiece focal length, not the objective's.
Why D is wrong: D is wrong because it divides L by 15.0 and then by a further factor, compounding the same omission of f_e from the tube sum.
MCQ 8Direct ApplicationPractice
A refracting astronomical telescope in normal adjustment has an objective of focal length 2.10 m and a magnifying power of magnitude 70.0. The focal length of its eyepiece is
Show answer and why every option is right or wrong
Answer: A. A is correct. Rearranging M = f_o/f_e for the eyepiece focal length gives f_e = f_o/M = 2.10/70.0 = 3.00 × 10⁻² m — NCERT Class 12 Physics Part 2, Chapter 9, page 244.
Why B is wrong: B gives 3.33 × 10¹ m (33.3 m) — this comes from evaluating M/f_o = 70.0/2.10 = 33.3 instead of f_o/M. The magnification relation is M = f_o/f_e, so solving for f_e means dividing f_o by M, not M by f_o. (trap: inverts f_o/f_e ratio)
Why C is wrong: C gives 2.10 × 10⁰ m (2.10 m) — this simply restates the objective's own focal length without dividing by M at all, as if f_e = f_o. A telescope's eyepiece must be far shorter than its objective to give real magnification.
Why D is wrong: D gives 1.47 × 10² m (147 m) — this multiplies f_o by M (2.10 × 70.0) instead of dividing. Multiplying, rather than dividing, cannot shrink a large objective focal length down to the short eyepiece length the design requires.
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Astronomical Telescope: quick recall before you leave
How do you solve a Astronomical Telescope question? A worked example
Pattern: telescope magnification and tube length (highest in-scope relevance: 3 occurrences, 2020/2024/2025).
- 1
Given
Objective focal length f_o = 1.44 m
Eyepiece focal length f_e = 1.20 × 10⁻² m
Telescope in normal adjustment; object at infinity. - 2
Required
The magnifying power (with sign) and the tube length.
- 3
Concept
The objective forms a real inverted image of the distant object at its focal plane. In normal adjustment the eyepiece is positioned so this image lies at its focal plane too, sending the final image to infinity. Hence the two lenses are separated by f_o + f_e, and the angular magnification is the ratio of the focal lengths.
- 4
Formula
M = −f_o/f_e and L = f_o + f_e
- 5
Substitution
M = −(1.44) / (1.20 × 10⁻²)
L = 1.44 + 1.20 × 10⁻² - 6
Calculation
M = −120.
L = 1.4520 m.
The integers in these relations (the ratio and the sum carry no numerical constants) are exact and impose no sig-fig limit; the precision is set entirely by the two measured focal lengths, each given to three significant figures. - 7
Final answer
M = −1.20 × 10², i.e. magnifying power 120 with an inverted final image.
L = 1.45 m. - 8
Common trap
Writing M = f_e/f_o gives 8.33 × 10⁻³ — a "magnification" smaller than 1 for an instrument built to magnify. The corpus records this inversion as the dominant distractor for this pattern. Before you compute, fix the physical direction: f_o is the big one, and it belongs on top. A second slip is reporting L = 1.44 m by dropping f_e from the sum; here f_e is only 1.2 cm, so the difference is small but the options are built to separate 1.44 from 1.45.
- 9
Similar NEET-style question
A telescope in normal adjustment has magnifying power of magnitude 25.0 and an objective of focal length 1.00 m. Find the eyepiece focal length and the tube length. *(f_e = 4.00 × 10⁻² m; L = 1.04 m.)*
What to remember before solving Astronomical Telescope questions
Which Astronomical Telescope formulas do you need for NEET?
1 formula — click to collapse
Astronomical telescope magnification (normal)
Magnification of astronomical telescope in normal adjustment (final image at infinity).
| Symbol | Quantity | SI Unit |
|---|---|---|
| f_o | objective focal length | m |
| f_e | eyepiece focal length | m |
Valid when
- Normal adjustment (image at infinity)
- Object at infinity
More in Optics: 9 exam traps and mistakes · 10 formulas · 6 question patterns from its other lessons.
Astronomical Telescope questions from past NEET papers
2 questions from NEET 2020, 2024. Answers verified against NTA official keys. — click to collapse
NEET 2024Revised key
134
228
317
432
NTA Answer: Option 2(revised_final)
NEET 2020
16.00×10⁻⁷ rad
23.66×10⁻⁷ rad
31.83×10⁻⁷ rad
47.32×10⁻⁷ rad
NTA Answer: Option 2(final)
How does NEET ask about Astronomical Telescope?
1 recurring pattern from past papers — click to collapse
Telescope magnification M = -f_o/f_e. Tube length L = f_o + f_e.
Direct ApplicationEasy
Common distractors
inverts fo fe
Uses f_e/f_o instead of f_o/f_e
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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