PhysicsISC Class 12 · Ray Optics and Optical Instruments

Microscopes and Telescopes: Ray Diagrams, Magnifying Power and Uses

Updated 15 October 2026 · Fact-checked

A simple microscope (one convex lens) has magnifying power $m = 1 + \frac{D}{f}$ with the image at D and $\frac{D}{f}$ with it at infinity. A compound microscope gives $m = m_o \times m_e$. An astronomical telescope in normal adjustment gives $m = \frac{f_o}{f_e}$ and has length $f_o + f_e$. D is 25 cm.

Understand Microscopes and Telescopes: Magnifying Power

Magnifying power (angular magnification) is the ratio of the angle subtended at the eye by the image to the angle subtended by the object when viewed directly in the most favourable way: at D = 25 cm for a microscope, and at its actual (very large) distance for a telescope. It is a pure number, not the "power" in dioptres.

Simple microscope (magnifying glass). A convex lens of short focal length f; the object is placed between the optical centre and F, giving a virtual, erect, magnified image on the same side as the object.

*Image at D*: put $v = - D$ in the lens formula ( $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ , Cartesian): $u = - \frac{f D}{f + D}$ . The angles are $β = \frac{h}{∣ u ∣}$ (with lens) and $α = \frac{h}{D}$ (unaided), so

$m = \frac{β}{α} = \frac{D}{∣ u ∣} = 1 + \frac{D}{f} .$

*Image at infinity* (object at F, relaxed eye): $m = \frac{D}{f}$ .

Compound microscope. Two convex lenses: an objective of very short focal length and small aperture, and an eyepiece of slightly larger focal length and aperture. The object is placed just beyond $F_{o}$ ; the objective forms a real, inverted, magnified image inside the tube, which lies within $f_{e}$ of the eyepiece; the eyepiece acts as a simple microscope on it. The final image is virtual and inverted with respect to the object.

*Final image at D*: $m = m_{o} m_{e} = \frac{v _{o}}{u _{o}} (1 + \frac{D}{f _{e}})$ (magnitude).
*Final image at infinity* (expression only): $m \approx \frac{L}{f _{o}} \cdot \frac{D}{f _{e}}$ , where L is the tube length (distance between the inner foci).
*Resolving power*: the least separation it can resolve is $d_{min} = \frac{1.22 λ}{2 n s i n β}$ (n = refractive index of the medium between object and objective, β = half-angle of the cone of light accepted). Resolving power $= \frac{1}{d _{min}} = \frac{2 n s i n β}{1.22 λ}$ : it improves with shorter wavelength and with oil immersion.

Astronomical (refracting) telescope. Objective: convex lens of large focal length and large aperture; eyepiece: convex lens of short focal length. Light from a distant star arrives parallel; the objective forms a real, inverted image at its focus.

*Normal adjustment* (final image at infinity): the image lies at $F_{e}$ , so $L = f_{o} + f_{e}$ . If the object subtends α and the final image β, $α = \frac{h}{f _{o}}$ , $β = \frac{h}{f _{e}}$ , so

$m = \frac{β}{α} = \frac{f _{o}}{f _{e}} .$

*Final image at D*: $m = \frac{f _{o}}{f _{e}} (1 + \frac{f _{e}}{D})$ and $L = f_{o} + ∣ u_{e} ∣$ , with $∣ u_{e} ∣ = \frac{f _{e} D}{f _{e} + D}$ .

Reflecting telescope (Newtonian / Cassegrain). A concave (preferably parabolic) mirror is the objective; a small plane or convex mirror sends the light to the eyepiece. Advantages: no chromatic aberration (reflection does not depend on wavelength); a parabolic mirror has no spherical aberration; a mirror can be very large and supported from behind, giving more light-gathering and better resolution; it is lighter and cheaper than a lens of equal aperture. Disadvantage: the secondary mirror blocks a little light. (This was asked in ISC 2025, Q1(B)(iv), about the Hubble telescope.)

Comparing the two instruments (ISC 2024 Q20). Microscope: objective has smaller focal length and aperture than the eyepiece. Telescope: objective has larger focal length and aperture than the eyepiece.

Microscopes and Telescopes: Magnifying Power: formulas and key facts

Simple microscope, image at D: $m = 1 + \frac{D}{f}$; D = 25 cm.
Simple microscope, image at infinity: $m = \frac{D}{f}$; Object at F.
Compound microscope, image at D: $m = \frac{v _{o}}{u _{o}} (1 + \frac{D}{f _{e}})$; Magnitude.
Compound microscope, image at infinity: $m \approx \frac{L}{f _{o}} \cdot \frac{D}{f _{e}}$; Expression only.
Telescope, normal adjustment: $m = \frac{f _{o}}{f _{e}}, L = f_{o} + f_{e}$; Image at infinity.
Telescope, image at D: $m = \frac{f _{o}}{f _{e}} (1 + \frac{f _{e}}{D})$; L = f_o + |u_e|.
Microscope resolving power: $\frac{1}{d _{min}} = \frac{2 n s i n β}{1.22 λ}$; Higher with smaller λ.

How to solve Microscopes and Telescopes: Magnifying Power questions

Decide the instrument and where the final image is; then pick the matching formula, or trace lens by lens.

1Note D = 25 cm and keep all lengths in cm (or all in m).
2Compound microscope: use the lens formula for the objective to get $v_{o}$ ; $m_{o} = \frac{v _{o}}{u _{o}}$ .
3Eyepiece: for image at D, $m_{e} = 1 + \frac{D}{f _{e}}$ and $∣ u_{e} ∣ = \frac{f _{e} D}{f _{e} + D}$ ; for infinity, $m_{e} = \frac{D}{f _{e}}$ , $∣ u_{e} ∣ = f_{e}$ .
4Tube length (lens separation) $= v_{o} + ∣ u_{e} ∣$ .
5Telescope: $m = \frac{f _{o}}{f _{e}}$ (normal) and $L = f_{o} + f_{e}$ ; for image at D multiply by $(1 + \frac{f _{e}}{D})$ .
6Give magnifying power as a number (no unit); state whether the image is erect or inverted if asked.

Microscopes and Telescopes: Magnifying Power in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A jeweller's loupe

Suppose a jeweller uses a loupe marked "10×", meaning a magnifying power of 10 for a relaxed eye (image at infinity), to inspect a diamond.

Find its focal length, and its magnifying power and the object distance if the jeweller focuses the image at 25 cm instead.

Try it first, then show the solutionSolution

$\frac{D}{f} = 10$ , so f = 2.5 cm.
Image at D: $m = 1 + \frac{25}{2.5} = 11$ .
Object distance $= \frac{f D}{f + D} = \frac{2.5 \times 25}{27.5} \approx 2.27$ cm, just inside F.

Answer: f = 2.5 cm; m = 11 with the diamond about 2.27 cm from the lens.

Memory hook: Relaxed eye: D/f. Strained eye at D: one more.

Seeing Saturn's rings

Imagine an amateur astronomer with a telescope of objective focal length 90 cm and eyepiece focal length 1.8 cm, looking at Saturn's rings, which span about 45 seconds of arc (about $2.18 \times 1 0^{- 4}$ rad).

Find the magnifying power in normal adjustment and the angle the rings subtend through the telescope.

Try it first, then show the solutionSolution

$m = \frac{90}{1.8} = 50$ .
Angle through the telescope $= 50 \times 2.18 \times 1 0^{- 4} \approx 0.0109$ rad ≈ 0.63°, larger than the Moon looks to the naked eye.

Answer: 50; about 0.63°, enough to see the ring shape clearly.

Memory hook: Telescope: big over small, fo over fe.

Tricks to remember Microscopes and Telescopes: Magnifying Power

Short memory aids. Say each one aloud once, then use it on a question the same day.

Microscope small first, telescope big first
Microscope objective: small f, small aperture. Telescope objective: big f, big aperture.
Use it when: Comparison questions (2024 Q20).
Plus one at D
Image at D adds "1 +" (microscope) or "1 + fe/D" (telescope).
Use it when: Choosing the formula.
Telescope length adds
Normal adjustment: $L = f_{o} + f_{e}$ .
Use it when: Tube length.
Magnifying power is not dioptres
m is a ratio of angles; P = 1/f is lens power.
Use it when: Avoiding the 2024 slip.

Microscopes and Telescopes: Magnifying Power: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Telescope in normal adjustment · 1 mark · MCQBoard level

The magnifying power of an astronomical telescope in normal adjustment is:
(a) $f_{o} + f_{e}$
(b) $f_{o} f_{e}$
(c) $\frac{f _{o}}{f _{e}}$
(d) $\frac{f _{e}}{f _{o}}$

Try it first, then show the solutionSolution

$α = \frac{h}{f _{o}}$ , $β = \frac{h}{f _{e}}$ , so $m = \frac{f _{o}}{f _{e}}$ .

Answer: (c).

Example 2: Magnifying glass · 2 marks · Short answerBoard level

A convex lens of focal length 4 cm is used as a simple microscope. Find its magnifying power with the image at 25 cm and at infinity, and the object distance in the first case.

Try it first, then show the solutionSolution

At D: $m = 1 + \frac{25}{4} = 7.25$ ; at infinity: $m = \frac{25}{4} = 6.25$ . (1)
Object distance $= \frac{f D}{f + D} = \frac{100}{29} \approx 3.45$ cm from the lens, inside F. (1)

Answer: 7.25; 6.25; ≈ 3.45 cm.

Example 3: Derivation: simple microscope at D · 3 marks · Long answerHarder board level

With a ray diagram, derive the magnifying power of a simple microscope when the final image is formed at the least distance of distinct vision.

Try it first, then show the solutionSolution

Diagram: object AB between O and F; rays give a virtual, erect, magnified image A′B′ at distance D on the same side. (1)
Unaided, the object at D subtends $α = \frac{h}{D}$ ; with the lens, the image subtends $β = \frac{h}{∣ u ∣}$ (same angle as the object at the lens). So $m = \frac{D}{∣ u ∣}$ . (1)
Lens formula with v = −D: $\frac{1}{- D} - \frac{1}{u} = \frac{1}{f}$ gives $\frac{D}{∣ u ∣} = 1 + \frac{D}{f}$ . Hence $m = 1 + \frac{D}{f}$ . (1)

Answer: $m = 1 + \frac{D}{f}$ .

Example 4: Compound microscope at D · 3 marks · Short answerHarder board level

A compound microscope has an objective of focal length 2 cm and an eyepiece of focal length 5 cm. An object is 2.5 cm from the objective, and the final image is at 25 cm. Find the magnifying power and the separation of the lenses.

Try it first, then show the solutionSolution

Objective: $\frac{1}{v _{o}} = \frac{1}{2} - \frac{1}{2.5} = 0.1$ , so $v_{o} = 10$ cm; $m_{o} = \frac{10}{- 2.5} = - 4$ . (1)
Eyepiece: $m_{e} = 1 + \frac{25}{5} = 6$ ; $m = - 24$ (24 times, inverted). (1)
$∣ u_{e} ∣ = \frac{5 \times 25}{30} \approx 4.17$ cm; separation $= 10 + 4.17 \approx 14.2$ cm. (1)

Answer: 24 (inverted); ≈ 14.2 cm.

Example 5: Telescope and the Moon · 4 marks · Long answerHOTS / competency

A telescope has $f_{o}$ = 150 cm and $f_{e}$ = 6 cm. (a) Find m and the tube length in normal adjustment. (b) Find m and the tube length with the final image at 25 cm. (c) The Moon (diameter $3.48 \times 1 0^{6}$ m, distance $3.84 \times 1 0^{8}$ m) is viewed. Find the diameter of the image formed by the objective.

Try it first, then show the solutionSolution

(a) $m = \frac{150}{6} = 25$ ; L = 156 cm. (1)
(b) $m = 25 (1 + \frac{6}{25}) = 31$ ; $∣ u_{e} ∣ = \frac{6 \times 25}{31} \approx 4.84$ cm, L ≈ 154.8 cm. (1)
(c) The Moon subtends $α = \frac{3.48 \times 1 0 ^{6}}{3.84 \times 1 0 ^{8}} \approx 9.06 \times 1 0^{- 3}$ rad. (1)
Image at the focal plane: diameter $= f_{o} α = 150 \times 9.06 \times 1 0^{- 3} \approx 1.36$ cm. (1)

Answer: (a) 25, 156 cm; (b) 31, ≈ 154.8 cm; (c) ≈ 1.36 cm.

Example 6: Building a reflecting telescope · 4 marks · Case-basedHOTS / competency

Suppose a school builds a reflecting telescope with a concave mirror of radius of curvature 4 m and an eyepiece of focal length 2.5 cm.
(i) Find the focal length of the mirror. (1 mark)
(ii) Find the magnifying power in normal adjustment. (1 mark)
(iii) Why is the mirror made parabolic rather than spherical? (1 mark)
(iv) Give two reasons why large telescopes use mirrors instead of lenses. (1 mark)

Try it first, then show the solutionSolution

(i) $f = \frac{R}{2} = 2$ m = 200 cm.
(ii) $m = \frac{200}{2.5} = 80$ .
(iii) A parabolic mirror brings all parallel rays to one focus, so there is no spherical aberration.
(iv) No chromatic aberration; a large mirror can be supported from behind, is lighter and cheaper, and its large aperture gathers more light and resolves better.

Answer: (i) 2 m; (ii) 80; (iii) removes spherical aberration; (iv) as listed.

Where students lose marks in Microscopes and Telescopes: Magnifying Power

The slips examiners see most often, and what to write instead.

Mistake: Writing "power" for magnifying power
The ISC 2024 examiners (Q20(iv)) saw "Power" instead of magnifying power or magnification.
Do this: Say "magnifying power" (angular magnification); lens power is in dioptres and different.
Usually costs: 1 mark.
Mistake: Comparing uses instead of lenses
Also in 2024 Q20(v): candidates gave uses or drew diagrams when asked to compare objective and eyepiece.
Do this: Compare one feature: focal length or aperture of objective versus eyepiece.
Usually costs: 1 mark.
Mistake: Wrong object position for a magnifying glass
In 2025 Q13 candidates placed the object at F or beyond F and called the image real.
Do this: Object between O and F; image virtual, erect, magnified, same side.
Usually costs: 1 to 2 marks.
Mistake: Giving a commercial reason for mirrors
In 2025 Q1(B)(iv) some candidates gave cost-based or vague answers.
Do this: Quote a scientific advantage: no chromatic aberration; larger aperture, brighter image.
Usually costs: 1 mark.

Exam tips

Draw ray diagrams with arrows, label F, objective, eyepiece and the intermediate image.
State where the final image is (D or infinity) before quoting a formula.
For a compound microscope, use the lens formula for the objective; do not assume $v_{o}$ = L.
Magnifying power has no unit; tube length does.

Practice questions on Microscopes and Telescopes: Magnifying Power

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
In a compound microscope:
(a) the objective has a larger focal length than the eyepiece
(b) the objective has a shorter focal length and smaller aperture than the eyepiece
(c) both lenses have the same focal length
(d) the objective is concave
Hint
Opposite of a telescope.
Answer
(b).
Q2 · 1 mark · Short answerBoard level
Find the magnifying power of a simple microscope of focal length 2 cm for a relaxed eye.
Hint
Image at infinity.
Answer
$\frac{25}{2} = 12.5$ .
Q3 · 2 marks · Short answerHarder board level
A telescope has $f_{o}$ = 120 cm and $f_{e}$ = 4 cm. Find its magnifying power and length in normal adjustment.
Hint
Ratio and sum.
Answer
30; 124 cm.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The objective of a reflecting telescope does not suffer from chromatic aberration.
Reason (R): The law of reflection does not depend on the wavelength of light.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Dispersion needs refraction.
Answer
(a).
Q5 · 2 marks · Short answerHarder board level
A compound microscope with $f_{o}$ = 0.8 cm and $f_{e}$ = 2.5 cm has tube length 16 cm. Find its magnifying power with the final image at infinity.
Hint
$\frac{L}{f _{o}} \cdot \frac{D}{f _{e}}$ .
Answer
20 × 10 = 200.
Q6 · 3 marks · Short answerHOTS / competency
A microscope uses oil of refractive index 1.5 and accepts a cone of half-angle β with sin β = 0.9. Find the least separation it can resolve with light of 500 nm. How does its resolving power change with 400 nm light?
Hint
$d_{min} = \frac{1.22 λ}{2 n s i n β}$ .
Answer
$d_{min} \approx 2.26 \times 1 0^{- 7}$ m; resolving power increases by $\frac{500}{400} = 1.25$ times.
Q7 · 2 marks · Short answerHOTS / competency
A telescope in normal adjustment is 84 cm long and has magnifying power 20. Find the focal lengths of its lenses.
Hint
$f_{o} + f_{e} = 84$ , $f_{o} = 20 f_{e}$ .
Answer
$f_{e}$ = 4 cm, $f_{o}$ = 80 cm.
Q8 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a hobby microscope has an objective of focal length 1.5 cm and an eyepiece of 3 cm. A specimen is placed 1.8 cm from the objective and the final image is at 25 cm.
(i) Find the image distance from the objective. (1 mark)
(ii) Find $m_{o}$ and $m_{e}$ . (1 mark)
(iii) Find the total magnifying power. (1 mark)
(iv) Find the distance between the lenses. (1 mark)
Hint
$\frac{1}{v} = \frac{1}{1.5} - \frac{1}{1.8}$ .
Answer
(i) 9 cm. (ii) $m_{o} = - 5$ , $m_{e} \approx 9.33$ . (iii) ≈ 46.7 (inverted). (iv) $∣ u_{e} ∣ \approx 2.68$ cm, separation ≈ 11.7 cm.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Simple microscope, image at D?Show answer
$m = 1 + \frac{D}{f}$ .
2.Telescope length in normal adjustment?Show answer
$f_{o} + f_{e}$ .
3. $f_{o}$ = 100 cm, $f_{e}$ = 2 cm: m (normal)?Show answer
50.
4.Which has the larger aperture in a telescope?Show answer
The objective.
5.One advantage of a reflecting telescope?Show answer
No chromatic aberration (also large aperture, no spherical aberration if parabolic).

Microscopes and Telescopes: Magnifying Power: frequently asked questions

Why is the objective of a telescope large?

A large aperture collects more light, giving a brighter image of faint objects, and improves resolving power.

Why does a microscope objective have a short focal length?

The magnification of the objective \frac{v_o}{u_o} is large when the object is just beyond a short focal length, and a short f_e adds more magnification.

Is the final image in an astronomical telescope erect?

No, it is inverted. This does not matter for stars and planets.

What does the ISC 2027 syllabus need here?

Ray diagrams and derivations for the simple microscope (image at D and infinity), the compound microscope (image at D; only the expression at infinity), refracting telescope (infinity and D), the reflecting telescope ray diagram, resolving power of the compound microscope, and advantages and uses.

How was this asked recently?

ISC 2024 Q20 (5 marks, passage-based) and ISC 2025 Q13 (3 marks, magnifying glass) and Q1(B)(iv) (1 mark, mirror objective).