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. 244Magnifying Powers
8 MCQs1 revision card9-step worked example
Source: NCERT OpticsOfficial key: NTA-verifiedLast updated: 27 Sep 2026
Magnifying Powers, explained for NEET
Magnifying power is an angular ratio, not a linear one. The commonest confusion in this topic is importing the linear magnification habit — m = −v/u, a height ratio — into instrument problems where the quantity asked is the ratio of the angle the image subtends at the eye to the angle the object subtends unaided. A telescope looking at the Moon produces an image far smaller than the Moon in linear terms and still magnifies, because only angles matter to the eye.
That angular definition is why the two instruments have different-looking formulas built the same way. NCERT Class 12 Physics Part 2, Chapter 9, page 244 gives the astronomical telescope in normal adjustment as the ratio of objective to eyepiece focal length, M = −f_o/f_e, with tube length L = f_o + f_e. The microscope's magnifying power is a product: the objective's linear magnification multiplied by the eyepiece's angular magnification. One is a ratio of two focal lengths; the other is a product of two stages.
Read the direction of the ratio carefully. For the telescope, large objective focal length over small eyepiece focal length — a long objective and a short eyepiece give high power. For the compound microscope the objective focal length is small and appears in the denominator of its stage. The two instruments pull the same symbol in opposite directions, and a question that gives you f_o and f_e without naming the instrument is testing exactly that.
The negative sign in M = −f_o/f_e records inversion of the final image, not a reduction. NEET stems usually ask for magnifying power and expect the magnitude; when the stem asks whether the image is erect or inverted, the sign is the answer.
Watch-out: high magnifying power and high resolving power are different claims. Increasing f_o/f_e does not let a telescope separate two stars that its aperture cannot resolve.
Can you answer these Magnifying Powers 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
The magnifying power of an optical instrument is defined as the ratio of
Show answer and why every option is right or wrong
Answer: C. Magnifying power is an angular quantity — the ratio of the visual angle with the instrument to the visual angle without it. NCERT Class 12 Physics Part 2, Chapter 9, page 244 defines the magnifying powers of both instruments this way.
Why A is wrong: A is wrong because that ratio is linear magnification, a height ratio used for a single mirror or lens; an instrument's magnifying power compares angles at the eye, which is why a telescope magnifies a Moon image far smaller than the Moon itself.
Why B is wrong: B is wrong because a ratio of distances is not a magnification of any kind; image and object distances enter the formulas only through the focal lengths.
Why D is wrong: D is wrong because f/D is the focal ratio, a measure of light-gathering and resolving ability, not of magnifying power.
MCQ 2Easy RecallPractice
In normal adjustment, the magnifying power of an astronomical telescope is given in magnitude by
Show answer and why every option is right or wrong
Answer: A. The astronomical telescope in normal adjustment has M = −f_o/f_e, objective focal length over eyepiece focal length. NCERT Class 12 Physics Part 2, Chapter 9, page 244.
Why B is wrong: B is wrong because f_o + f_e is the tube length L in normal adjustment, not the magnifying power — the two are separate results on the same page and are frequently swapped.
Why C is wrong: C is wrong because it inverts the ratio; f_e/f_o would make a short objective and a long eyepiece give high power, the opposite of how telescopes are built.
Why D is wrong: D is wrong because the product of two focal lengths has dimensions of area; magnifying power is a dimensionless ratio.
MCQ 3Easy RecallPractice
The magnifying power of a compound microscope is obtained by
Show answer and why every option is right or wrong
Answer: D. The compound microscope magnifies in two stages, so the total magnifying power is the product of the objective's linear magnification and the eyepiece's angular magnification. NCERT Class 12 Physics Part 2, Chapter 9, page 244.
Why A is wrong: A is wrong because the stages act in sequence on the same light, so their effects multiply; adding would give a value far too small for any real microscope.
Why B is wrong: B is wrong because discarding the objective stage would reduce the instrument to a simple magnifier.
Why C is wrong: C is wrong because f_o/f_e is the telescope result; the microscope's objective focal length is small and enters its own stage differently.
MCQ 4Direct ApplicationPractice
Two optical instruments are assembled from the same pair of lenses, one of focal length 1.00 m and one of focal length 2.0 × 10⁻² m. To obtain a high magnifying power in an astronomical telescope, the 1.00 m lens should be used as the
Show answer and why every option is right or wrong
Answer: D. M = f_o/f_e in magnitude, so high power needs the long-focus lens at the objective end and the short-focus lens at the eye end — here 1.00/0.020 = 50. NCERT Class 12 Physics Part 2, Chapter 9, page 244.
Why A is wrong: A is wrong because it places f_e in the numerator, inverting the ratio; this choice would give a magnifying power of 0.020, a reduction.
Why B is wrong: B is wrong because the tube length in normal adjustment is f_o + f_e, so it exceeds f_o; the reasoning stated does not support the assignment even though the assignment happens to be right.
Why C is wrong: C is wrong because it reaches the same inverted assignment as A by a different route, and the final image position in normal adjustment is at infinity regardless.
MCQ 5Direct ApplicationPractice
An astronomical telescope in normal adjustment has an objective of focal length 1.60 m and an eyepiece of focal length 4.0 × 10⁻² m. Its magnifying power in magnitude is
Show answer and why every option is right or wrong
Answer: B. M = f_o/f_e = 1.60/0.040 = 40 in magnitude. NCERT Class 12 Physics Part 2, Chapter 9, page 244.
Why A is wrong: A is wrong because it is f_e/f_o = 0.040/1.60, the inverted ratio — the single most common slip on this pattern.
Why C is wrong: C is wrong because 1.60 + 0.040 m is the tube length in metres, not the magnifying power.
Why D is wrong: D is wrong because it is the product f_o × f_e, which is dimensionally an area and cannot be a magnification.
MCQ 6Direct ApplicationPractice
A compound microscope has an objective that produces a linear magnification of 25.0 and an eyepiece whose angular magnification is 8.00. The magnifying power of the microscope is
Show answer and why every option is right or wrong
Answer: C. Microscope magnifying power is the product of the two stages: 25.0 × 8.00 = 200. NCERT Class 12 Physics Part 2, Chapter 9, page 244.
Why A is wrong: A is wrong because it divides the two stage magnifications, 25.0/8.00, importing the telescope's ratio structure into a microscope.
Why B is wrong: B is wrong because it adds the stages, 25.0 + 8.00; the stages act in sequence on the same light and therefore multiply.
Why D is wrong: D is wrong because it multiplies by a further factor of 10, not by 8.00.
MCQ 7Concept TrapPractice
A student reports that an astronomical telescope in normal adjustment has a magnifying power of −25.0. The negative sign indicates that
Show answer and why every option is right or wrong
Answer: B. In M = −f_o/f_e the negative sign records inversion of the final image; the magnitude 25.0 is the magnifying power the stem asks about elsewhere. NCERT Class 12 Physics Part 2, Chapter 9, page 244.
Why A is wrong: A is wrong because a magnitude of 25.0 means the image subtends 25.0 times the unaided visual angle; the sign carries orientation, not size.
Why C is wrong: C is wrong because the sign in this formula does not encode real or virtual; the final image in normal adjustment is at infinity and the sign would still be negative.
Why D is wrong: D is wrong because both focal lengths of a converging objective and eyepiece are positive; the minus sign is written into the formula itself, not produced by a negative input.
MCQ 8CalculationPractice
An astronomical telescope in normal adjustment has a tube length of 8.4 × 10⁻¹ m and a magnifying power of magnitude 20.0. The focal length of its eyepiece is
Show answer and why every option is right or wrong
Answer: A. With f_o = 20.0 f_e and f_o + f_e = 0.84 m, 21.0 f_e = 0.84 m, so f_e = 4.0 × 10⁻² m. NCERT Class 12 Physics Part 2, Chapter 9, page 244.
Why B is wrong: B is wrong because it divides the tube length by 20.0 instead of 21.0 — it forgets that the tube length contains the eyepiece focal length as well as the objective's.
Why C is wrong: C is wrong because it is the objective focal length f_o = 20.0 × 0.040 m, the other lens of the pair.
Why D is wrong: D is wrong because it multiplies the tube length by 20.0 instead of dividing, and exceeds the tube length itself, which no component focal length in normal adjustment can do.
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Magnifying Powers: quick recall before you leave
How do you solve a Magnifying Powers question? A worked example
- 1
Given
Astronomical telescope in normal adjustment.
Tube lengthL = 1.05 m.
Magnifying power magnitude|M| = 20.0. - 2
Required
The focal length of the objective,
f_o. - 3
Concept
In normal adjustment the object is at infinity and the final image is at infinity, so the intermediate image formed by the objective sits at the common focal point of the two lenses. That single geometric fact gives both relations at once: the magnifying power is the ratio of the focal lengths, and the tube length is their sum. Two equations, two unknowns.
- 4
Formula
|M| = f_o/f_eandL = f_o + f_e. - 5
Substitution
From the first relation,
f_o = 20.0 f_e.
Substituting into the second:20.0 f_e + f_e = 1.05 m. - 6
Calculation
21.0 f_e = 1.05 mf_e = 1.05/21.0 = 5.00 × 10⁻² mf_o = 20.0 × 5.00 × 10⁻² = 1.00 m
The 20.0 here is a measured magnifying power, and 1.05 m is a measured length — both carry three significant figures. The 21.0 is not a measurement: it arises as 20.0 + 1, where the 1 is the exact counting coefficient off_einL = f_o + f_e, so it contributes no sig-fig limit of its own. - 7
Final answer
f_o = 1.00 m(andf_e = 5.00 × 10⁻² m). - 8
Common trap
Dividing the tube length straight by the magnifying power —
1.05/20.0 = 5.25 × 10⁻² m— and reporting that asf_e. The tube length isf_o + f_e, notf_oalone, so the divisor is|M| + 1, never|M|. The error is small enough to look plausible and is routinely offered as an option. The other direction of the same slip is inverting the ratio and solving for the wrong lens: check at the end that the objective came out the longer of the two, since that is what makes the instrument magnify. - 9
Similar NEET-style question
An astronomical telescope in normal adjustment has an objective of focal length 0.96 m and a tube length of 1.00 m. Find its magnifying power in magnitude. *(Work backwards:
f_e = L − f_o, then|M| = f_o/f_e.)*
What to remember before solving Magnifying Powers questions
Which Magnifying Powers 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.
Magnifying Powers questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
How does NEET ask about Magnifying Powers?
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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