Optical Instruments

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Optical instruments use reflection, refraction and diffraction of light to produce images for observation, measurement or analysis. Three instruments dominate the NABTEB syllabus: the simple microscope, the compound microscope, and the astronomical telescope, alongside the human eye as a near-point optical system (D = 25 cm).

Instrument	Magnifying power (M)	Where used
Simple microscope	M = D/f + 1	Reading small print
Compound microscope	M = (L/f₀) × (D/fₑ)	Biology/biology lab specimens
Astronomical telescope (normal adjustment)	M = f₀/fₑ	Viewing distant celestial bodies

Exam tip: For NABTEB, expect one 2-mark calculation asking you to substitute f, L and D into the correct formula, then a 6-mark ray diagram of either the microscope or telescope in normal adjustment. Memorise the sign convention (real objects: u negative; real images: v positive for a converging lens).

Key quantities — focal length f in metres, power P = 1/f in dioptres (D), and the lens equation 1/f = 1/u + 1/v — appear in almost every objective question on this topic.

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Lens formula and sign convention

The thin-lens equation 1/f = 1/u + 1/v (with f, u, v all in metres) governs image position for both converging and diverging lenses. Using the real-is-positive convention adopted by NABTEB, an object placed in front of a converging lens gives u positive, the image distance v positive if it forms on the opposite side, and f positive. For a diverging lens, f is negative. Linear magnification M = v/u is dimensionless.

Simple microscope

A single converging lens of short focal length f is held so the object lies just inside f. The eye views a virtual, magnified, upright image at the least distance of distinct vision D = 0.25 m.

M = D/f + 1
The “+1” appears because the relaxed-eye position is infinity, not D — examiners love testing whether students drop it.

Compound microscope

Two converging lenses combine: a short-focal-length objective (f₀) forms a real, magnified intermediate image at its focal plane, and an eyepiece (fₑ) acting as a simple microscope re-magnifies it.

M = (L/f₀) × (D/fₑ)
where L is the tube length (distance between the two lenses’ principal planes, ≈ 0.16–0.20 m in a school lab).

Astronomical telescope in normal adjustment

The objective (f₀, large) collects light from a distant object, focusing it at its focal plane; the eyepiece (fₑ, smaller) is positioned so its focal point coincides with the objective’s focal point. The final image is at infinity, so the eye is relaxed.

M = f₀/fₑ

Defects of vision and their correction

Myopia (short-sightedness): far point closer than infinity → corrected by a concave/diverging lens of focal length equal to the far-point distance.
Hypermetropia (long-sightedness): near point farther than 25 cm → corrected by a convex/converging lens of suitable power P = 1/f.
Presbyopia: loss of accommodation with age → corrected by bifocals (convex for near, concave for far).

Defect	Symptom	Lens used	Power sign
Myopia	Distant objects blurred	Concave	Negative
Hypermetropia	Near objects blurred	Convex	Positive
Presbyopia	Both distances affected	Bifocal	Combination

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Resolving power and Rayleigh’s criterion

An instrument’s resolving power is its ability to render two close point sources as separate images. Rayleigh’s criterion states that the images are just resolved when the central maximum of one diffraction pattern coincides with the first minimum of the other. For a microscope objective, resolving power improves with larger numerical aperture and shorter wavelength of illumination — which is why blue light and oil-immersion objectives yield finer detail.

Worked example (NABTEB-style, 4 marks)

A compound microscope has an objective of focal length 1.5 cm, an eyepiece of focal length 5.0 cm, and the final virtual image is formed at D = 25 cm from the eye. The tube length L is 20 cm. Find the magnifying power.

Using M = (L/f₀) × (D/fₑ):
L/f₀ = 20 / 1.5 ≈ 13.33; D/fₑ = 25 / 5.0 = 5.0.
Therefore M = 13.33 × 5.0 ≈ 66.7× (to 3 s.f.).
Common error: students forget to convert cm to m, then divide — examiners award 1 mark for each correct substitution and 1 for the final numerical answer with units of dimensionless magnification.

Telescope edge cases

In non-normal adjustment, the final image is formed at D instead of infinity, and the formula becomes M = (f₀/fₑ) + (f₀·D)/(fₑ·L). NABTEB occasionally tests this as a 3-mark extension. A terrestrial telescope inserts an erecting lens between objective and eyepiece to invert the inverted image from the astronomical telescope.

Connection to wave optics

Young’s double-slit fringes (spacing s = λD/d) and the grating equation d sin θ = nλ are tested in the same paper. Candidates confuse grating spacing d (in metres) with slit width — remember: d = 1/N, where N is lines per metre of the grating.

Common mistakes

Using M = v/u for a microscope or telescope instead of the angular form.
Dropping the +1 in the simple-microscope formula.
Treating f negative for a converging lens (it is positive under the real-is-positive convention).
Stating a telescope’s power without specifying normal adjustment.

Practice prompts

A myopic patient reads a book held at 20 cm and cannot see objects beyond 2 m. Determine the focal length and power of the correcting lens.
A telescope objective has f₀ = 100 cm and eyepiece fₑ = 5 cm. Find the magnifying power in normal adjustment, and the tube length.

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Optical Instruments