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Optical instruments questions look like formula substitution, which is why students lose marks on them: the formula depends on where the final image is, and the wrong choice gives a confident wrong answer. This post builds both instruments from the lens formula, works one full example for each, and lists the slips to watch for.
What do the syllabi cover?
The CBSE physics curriculum for 2025-26 lists "microscopes and astronomical telescopes (reflecting and refracting) and their magnifying powers". The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus use almost the same words. The CBSE 2025-26 sample paper includes a three-mark compound microscope question with the final image at the near point, so expect both cases.
How does a simple magnifier work?
The NCERT chapter on ray optics and optical instruments starts with one convex lens of focal length f and takes D ≈ 25 cm as the near point.
Final image at the near point: m = 1 + D/f.
Final image at infinity (relaxed eye): m = D/f.
The near-point value is larger by exactly 1, but viewing at infinity is more comfortable. NCERT notes that a single lens gives a magnification of at most about 9 for realistic focal lengths. For more, you need two lenses.
How does a compound microscope get its magnification?
The objective (short focal length fₒ) forms a real, inverted, magnified image near the eyepiece. The eyepiece (focal length fₑ) acts as a simple magnifier on that image. Total magnification is the product:
Objective: mₒ = v/u for the objective, which NCERT approximates as L/fₒ, where L is the tube length (distance between the objective's second focal point and the eyepiece's first focal point).
Eyepiece: mₑ = 1 + D/fₑ (near point) or D/fₑ (infinity).
So m = (L/fₒ)(D/fₑ) for the final image at infinity.
Worked example. A microscope has fₒ = 1.0 cm and fₑ = 2.5 cm. An object is 1.1 cm in front of the objective, and the final image is at the near point, 25 cm from the eye. Find the magnifying power and the distance between the lenses.
Objective image: use 1/v − 1/u = 1/f with u = −1.1 cm. 1/v = 1/1.0 − 1/1.1 = 0.0909, so v = +11 cm.
Objective magnification: |mₒ| = v/|u| = 11/1.1 = 10.
Eyepiece magnification (near point): mₑ = 1 + 25/2.5 = 11.
Total: m = 10 × 11 = 110.
Eyepiece object distance: the final image is virtual at v = −25 cm. 1/u = 1/v − 1/f = −1/25 − 1/2.5 = −0.44, so u = −2.27 cm.
Separation: 11 + 2.27 ≈ 13.3 cm.
If the final image were at infinity instead, mₑ = 25/2.5 = 10, the eyepiece object would sit at its focal point, 2.5 cm away, and the total would be 10 × 10 = 100.
How does an astronomical telescope differ?
A telescope looks at distant objects, so the objective has a long focal length and a large aperture, and the eyepiece a short one. NCERT defines magnifying power as the ratio of the angle the final image makes at the eye to the angle the object makes. In normal adjustment (final image at infinity):
m = fₒ/fₑ, and the tube length is fₒ + fₑ.
With the final image at the near point, m = (fₒ/fₑ)(1 + fₑ/D).
Worked example. fₒ = 100 cm, fₑ = 5.0 cm.
Normal adjustment: m = 100/5 = 20; tube length = 105 cm.
Near point: m = 20 × (1 + 5/25) = 20 × 1.2 = 24.
Tube length for near point: for the eyepiece, v = −25 cm and f = 5 cm, so 1/u = −1/25 − 1/5 = −0.24 and u = −4.17 cm. Tube length = 100 + 4.17 ≈ 104.2 cm.
NCERT adds two facts worth a one-mark answer. Light-gathering power depends on the objective's area, and resolving power also improves with a larger objective. Large telescopes use mirrors because a mirror has no chromatic aberration and can be supported across its whole back. NCERT names the 2.34 m Cassegrain reflector at Kavalur, Tamil Nadu, as the largest telescope in India.
Common mistakes and how to avoid them
Swapping the focal lengths. A microscope needs both lenses short; a telescope needs a long objective and a short eyepiece.
Using the near-point formula for normal adjustment, or the reverse. Read the question for "relaxed eye", "infinity" or "least distance of distinct vision".
Dropping signs in the lens formula. Real images from the objective have positive v in the Cartesian convention; the virtual final image has negative v.
Calling angular magnification "size". A telescope makes the Moon look larger in angle, not bigger than it is.
Forgetting the tube length. Telescopes ask for fₒ + fₑ; microscopes ask for v of the objective plus |u| of the eyepiece.
Learners often carry their own ideas about images and vision through a course: Galili and Hazan (2000) studied high school and teacher-training students' knowledge of light and vision before and after ordinary instruction, and mapped the alternative ideas that persist. Drawing the ray diagram, not just the formula, is the fix. Our guide to drawing the diagram first helps with that step.
If you set up a problem and keep getting a different answer from the key, the Fermi Problem Library has a step-by-step solver that explains the reasoning, so you can find which line went wrong.
A practice routine for this chapter
Day 1: derive m = 1 + D/f and m = D/f for a simple magnifier from the lens formula.
Day 2: four compound microscope problems, two at the near point and two at infinity, each with lens separation.
Day 3: four telescope problems, including one asking for tube length at the near point.
Day 4: draw labelled ray diagrams for both instruments in normal adjustment.
Day 5: attempt the microscope question in the latest CBSE sample paper in eight minutes.
Today, redo the microscope example with the object at 1.2 cm and check that the magnification falls.
