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. 244Microscope
8 MCQs1 revision card9-step worked example
Source: NCERT OpticsPYQ coverage: NEET 2025Official key: NTA-verifiedLast updated: 27 Sep 2026
Microscope, explained for NEET
A compound microscope is two converging lenses doing two different jobs, and the usual error is treating them as one. The objective has a short focal length and sits close to the specimen; the eyepiece is a simple magnifier held at the eye. Reverse those roles and every downstream number is wrong.
The construction rule fixes everything else. The object is placed just beyond the objective's focal point, so the objective forms a real, inverted, enlarged image inside the tube. That intermediate image must land just inside the eyepiece's focal length — that is the only way the eyepiece can act as a magnifier and throw a virtual final image out at the near point or at infinity. The final image a viewer sees is virtual and inverted with respect to the original object.
Because both lenses converge and both contribute, magnification is a product, not a sum: the objective's linear magnification times the eyepiece's angular magnification. NCERT Class 12 Physics, Chapter 9, page 244 sets this out for both lens systems.
Two consequences worth holding. First, short focal lengths give large magnification here, which is the opposite of the telescope, where a long objective is what you want. A microscope objective of a few millimetres is normal. Second, the tube length is roughly the separation of the two lenses, and since the intermediate image must fall between them, the tube cannot be shorter than the objective's image distance.
One more distinction a NEET stem will lean on: the specimen is near, brightly illuminated, and its linear size is what you are enlarging. Nothing about a microscope involves an object at infinity — that assumption belongs to a different instrument in this chapter, and importing it is the single easiest way to lose a mark here.
Watch out for stems that hand you a focal length in centimetres and ask for an answer in millimetres, or that quietly call the eyepiece "the lens nearer the object."
Can you answer these Microscope 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 a compound microscope, the lens placed nearer to the object being examined is called the
Show answer and why every option is right or wrong
Answer: D. The lens facing the specimen is the objective, and in a compound microscope it is the shorter-focal-length lens of the pair — NCERT Class 12 Physics, Chapter 9, page 243.
Why A is wrong: A is wrong because the eyepiece is the lens at the viewing end, not the one nearest the object; the naming is reversed here.
Why B is wrong: B is wrong because although it gets the short focal length right for a microscope objective, it attaches the name 'eyepiece' to the lens nearest the object.
Why C is wrong: C is wrong because it names the lens correctly but gives it the longer focal length; the long-objective arrangement belongs to a telescope, not a microscope.
MCQ 2Easy RecallPractice
The image formed by the objective of a compound microscope is
Show answer and why every option is right or wrong
Answer: A. With the object just beyond the objective's focal point, the objective forms a real, inverted and enlarged intermediate image inside the tube — NCERT Class 12 Physics, Chapter 9, page 243.
Why B is wrong: B is wrong because a virtual erect diminished image would mean the object lay inside the focal length of a diverging arrangement; the objective is converging with the object outside its focus.
Why C is wrong: C is wrong because the intermediate image must be real — it has to be physically formed inside the tube for the eyepiece to use it as its object.
Why D is wrong: D is wrong because a converging lens with the object just outside the focal point enlarges and inverts; it does not produce an erect diminished image.
MCQ 3Easy RecallPractice
For the eyepiece of a compound microscope to act as a simple magnifier, the intermediate image formed by the objective must lie
Show answer and why every option is right or wrong
Answer: A. The eyepiece works as a magnifying glass only when its object — here the objective's intermediate image — lies just inside its focal length, producing an enlarged virtual final image. NCERT Class 12 Physics, Chapter 9, page 243.
Why B is wrong: B is wrong because the objective's focal point is where a distant object would image; the intermediate image forms well beyond it and its position relative to the eyepiece is what matters.
Why C is wrong: C is wrong because an object beyond twice the eyepiece's focal length would give a real diminished image, not the enlarged virtual image a magnifier produces.
Why D is wrong: D is wrong because an image behind the eyepiece would make it a virtual object; in the standard compound microscope the intermediate image forms in front of the eyepiece, inside the tube.
MCQ 4Direct ApplicationPractice
Two thin converging lenses, one of focal length 5.0 × 10⁻³ m and one of focal length 5.0 × 10⁻² m, are to be assembled into a compound microscope. To obtain the intended magnification, the lens of focal length 5.0 × 10⁻³ m should be used as the
Show answer and why every option is right or wrong
Answer: B. In a compound microscope the shorter-focal-length lens is the objective; placing the specimen just beyond its short focal point is what produces the large linear magnification of the intermediate image. NCERT Class 12 Physics, Chapter 9, page 243.
Why A is wrong: A is wrong because it inverts the microscope convention — the objective, not the eyepiece, carries the shorter focal length here.
Why C is wrong: C is wrong because eye comfort does not set the roles; the assignment follows from which lens must image the nearby specimen at high linear magnification.
Why D is wrong: D is wrong because although the total magnification is a product of two factors, each factor depends on which lens plays which role, so swapping the lenses changes the result.
MCQ 5Concept TrapPractice
In a compound microscope, the total magnification is obtained from the two lenses by
Show answer and why every option is right or wrong
Answer: B. The eyepiece magnifies an image that the objective has already enlarged, so the two factors compound: the total is their product. NCERT Class 12 Physics, Chapter 9, page 244.
Why A is wrong: A is wrong because adding would make a stage of magnification contribute a fixed amount rather than a factor; two stages of ×10 give ×100, not ×20.
Why C is wrong: C is wrong because a ratio would let a strong objective reduce the total, which contradicts the fact that both lenses enlarge.
Why D is wrong: D is wrong because the eyepiece's object is already the objective's enlarged image; discarding the objective's contribution ignores the first stage of enlargement entirely.
MCQ 6Direct ApplicationPractice
In a compound microscope, the objective produces a linear magnification of magnitude 40.0 and the eyepiece a further angular magnification of magnitude 5.00. The magnitude of the total magnification is
Show answer and why every option is right or wrong
Answer: D. The stages compound, so the magnitude of the total is 40.0 × 5.00 = 2.00 × 10². NCERT Class 12 Physics, Chapter 9, page 244.
Why A is wrong: A is wrong because it adds the two magnifications (40.0 + 5.00) instead of multiplying them.
Why B is wrong: B is wrong because it divides the objective's magnification by the eyepiece's; the two stages compound, they do not cancel.
Why C is wrong: C is wrong because it subtracts the eyepiece's contribution from the objective's, which would mean the second lens shrinks the image.
MCQ 7Direct ApplicationPractice
A compound microscope is required to give a total magnification of magnitude 6.00 × 10². Its eyepiece provides an angular magnification of magnitude 1.50 × 10¹. The magnitude of the linear magnification the objective must produce is
Show answer and why every option is right or wrong
Answer: C. Since the total is the product of the two stages, the objective must supply 6.00 × 10² ÷ 1.50 × 10¹ = 4.00 × 10¹. NCERT Class 12 Physics, Chapter 9, page 244.
Why A is wrong: A is wrong because it subtracts the eyepiece magnification from the total, applying an additive model to a multiplicative chain.
Why B is wrong: B is wrong because it multiplies the total by the eyepiece magnification instead of dividing; that treats the required total as though it were only the objective's share.
Why D is wrong: D is wrong because it adds the eyepiece magnification to the total, again treating the two stages as though they combined by addition.
MCQ 8CalculationPractice
A microscope objective of focal length 8.00 × 10⁻³ m is used with a specimen placed so that the objective forms its real image at a distance of 1.60 × 10⁻¹ m from the objective. The eyepiece has focal length 2.50 × 10⁻² m. Taking the objective's linear magnification as the ratio of its image distance to its object distance, and the eyepiece's angular magnification for a final image at infinity as the ratio of the near-point distance 2.50 × 10⁻¹ m to its focal length, the magnitude of the total magnification is closest to
Show answer and why every option is right or wrong
Answer: C. Using the thin-lens relation with f = 8.00 × 10⁻³ m and image distance 1.60 × 10⁻¹ m, the object sits 8.42 × 10⁻³ m from the objective, giving an objective magnification of magnitude 19.0; the eyepiece contributes 2.50 × 10⁻¹ ÷ 2.50 × 10⁻² = 10.0, and the product is 1.90 × 10². NCERT Class 12 Physics, Chapter 9, page 244.
Why A is wrong: A is wrong because it stops after the objective stage and never applies the eyepiece's factor of 10.0.
Why B is wrong: B is wrong because 2.09 × 10² is 19.0 × 11.0: it uses the near-point eyepiece factor 1 + D/f_e = 11.0 although the stem specifies a final image at infinity, where the eyepiece contributes D/f_e = 10.0, giving 19.0 × 10.0 = 1.90 × 10².
Why D is wrong: D is wrong because it uses only the eyepiece's contribution of 10.0 and then squares it, rather than pairing it with the objective's own magnification.
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Microscope: quick recall before you leave
How do you solve a Microscope question? A worked example
- 1
Given
Objective focal length f_o = 6.00 × 10⁻³ m.
Objective forms its real image at distance v_o = 1.20 × 10⁻¹ m from the objective.
Eyepiece focal length f_e = 5.00 × 10⁻² m.
Near-point distance D = 2.50 × 10⁻¹ m (exact, by convention). - 2
Required
The magnitude of the total magnification, with the final image at infinity.
- 3
Concept
A compound microscope magnifies in two stages. The objective forms a real enlarged image inside the tube; the eyepiece then views that image as a simple magnifier. The two stages compound, so the total is their product. The eyepiece's angular magnification with the final image at infinity is D/f_e.
- 4
Formula
Objective linear magnification: m_o = v_o / u_o (magnitude).
Object distance from the thin-lens relation with the objective.
Eyepiece angular magnification (final image at infinity): m_e = D / f_e.
Total: m = m_o × m_e. - 5
Substitution
For the objective, using magnitudes: 1/u_o = 1/f_o − 1/v_o = 1/(6.00 × 10⁻³) − 1/(1.20 × 10⁻¹).
For the eyepiece: m_e = (2.50 × 10⁻¹) / (5.00 × 10⁻²). - 6
Calculation
1/u_o = 166.67 − 8.333 = 158.33 m⁻¹, so u_o = 6.316 × 10⁻³ m.
m_o = (1.20 × 10⁻¹) / (6.316 × 10⁻³) = 19.00.
m_e = 5.00.
m = 19.00 × 5.00 = 95.0.
The near-point distance D = 2.50 × 10⁻¹ m is an exact conventional value, not a measurement, so it does not limit the significant-figure count. The three significant figures in the answer come from the three-figure focal lengths and image distance. - 7
Final answer
Total magnification of magnitude 9.50 × 10¹.
- 8
Common trap
Reading the 1.20 × 10⁻¹ m as the object distance rather than the image distance. It is the objective's image distance — the intermediate image sits that far down the tube. Feeding it in as u_o gives an objective magnification below 1, i.e. a microscope that shrinks, which should immediately flag the error. A second trap is approximating u_o ≈ f_o because the object is "close to the focus"; here that shortcut gives 20.0 rather than 19.00, enough to select a wrong option.
- 9
Similar NEET-style question
A compound microscope has an objective of focal length 5.00 × 10⁻³ m which forms its real image 1.00 × 10⁻¹ m from the objective, and an eyepiece of focal length 2.00 × 10⁻² m used with the final image at infinity. Taking the near-point distance as 2.50 × 10⁻¹ m (exact), find the magnitude of the total magnification.
What to remember before solving Microscope questions
More in Optics: 9 exam traps and mistakes · 11 formulas · 7 question patterns from its other lessons.
Microscope questions from past NEET papers
1 question from NEET 2025. Answers verified against NTA official keys. — click to collapse
NEET 2025
1250
2100
3125
4150
NTA Answer: Option 3(final)
Sources
NCERT refs: Class 12 Physics Chapter 9, p.244
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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