PhysicsCBSE Class 12 · Ray Optics and Optical Instruments

Microscopes and Telescopes Class 12: Ray Diagrams and Magnifying Power

Updated 15 October 2026 · Fact-checked

A compound microscope uses a short-focus objective to form a magnified real image that a short-focus eyepiece magnifies further: m = (L/f₀)(D/fₑ) in normal adjustment. An astronomical telescope uses a long-focus objective and short-focus eyepiece: m = f₀/fₑ, with tube length f₀ + fₑ. Reflecting telescopes use a concave mirror as the objective.

Understand Microscopes and Telescopes

Our eyes see an object larger when it subtends a larger angle. Optical instruments increase this angle. A microscope does it for tiny objects close at hand; a telescope does it for large objects far away. Magnifying power is the ratio of the angle subtended at the eye by the final image to the angle subtended by the object seen directly (at the near point D ≈ 25 cm for a microscope, or at its actual distance for a telescope).

Simple microscope (magnifying glass). A convex lens with the object inside its focal length gives a virtual, erect, enlarged image. If the image is at the near point, $m = 1 + \frac{D}{f}$ ; if it is at infinity (relaxed eye), $m = \frac{D}{f}$ . A shorter focal length gives more magnification, but very short focal lengths cause blurred images, which limits a single lens.

Compound microscope. Two converging lenses: an objective of very short focal length $f_{o}$ close to the object, and an eyepiece of short focal length $f_{e}$ .

The object is placed just beyond $F_{o}$ . The objective forms a real, inverted, magnified image near the eyepiece; its magnification is $m_{o} = \frac{v _{o}}{u _{o}} \approx \frac{L}{f _{o}}$ , where the tube length L is the distance between the second focal point of the objective and the first focal point of the eyepiece.
The eyepiece acts as a simple microscope on this image. In normal adjustment (final image at infinity), $m_{e} = \frac{D}{f _{e}}$ ; with the final image at the near point, $m_{e} = 1 + \frac{D}{f _{e}}$ .
Total: $m = m_{o} m_{e} \approx \frac{L}{f _{o}} \cdot \frac{D}{f _{e}}$ . The final image is inverted relative to the object.

Both focal lengths are kept small to make m large.

Astronomical (refracting) telescope. An objective of long focal length and large aperture and an eyepiece of short focal length.

Light from a distant object (angle α) forms a real, inverted image at the focal plane of the objective.
In normal adjustment, this image lies at the focus of the eyepiece too, so the final image is at infinity; the tube length is $f_{o} + f_{e}$ .
The image height is $f_{o} α$ and it subtends $β = \frac{f _{o} α}{f _{e}}$ at the eye, so $m = \frac{β}{α} = \frac{f _{o}}{f _{e}}$ .
If the final image is at the near point: $m = \frac{f _{o}}{f _{e}} (1 + \frac{f _{e}}{D})$ .

The large aperture gathers more light (to see faint objects) and gives finer detail.

Reflecting telescope. The objective is a large concave (paraboloidal) mirror. In the Cassegrain design, a small convex secondary mirror reflects the converging light back through a hole in the primary to the eyepiece. Advantages over refractors: no chromatic aberration (a mirror reflects all colours alike), reduced spherical aberration with a paraboloidal shape, and a mirror is lighter and can be supported over its whole back, so very large apertures are practical (large lenses are heavy, sag and are hard to make free of flaws).

Microscopes and Telescopes: formulas and key facts

Simple microscope: $m = 1 + \frac{D}{f} (image at D), m = \frac{D}{f} (image at \infty)$; D ≈ 25 cm.
Compound microscope (normal adjustment): $m = m_{o} m_{e} \approx \frac{L}{f _{o}} \cdot \frac{D}{f _{e}}$; L = tube length between the focal points.
Compound microscope (image at D): $m = \frac{v _{o}}{u _{o}} (1 + \frac{D}{f _{e}})$; Magnitude; the final image is inverted.
Astronomical telescope (normal adjustment): $m = \frac{f _{o}}{f _{e}}, L = f_{o} + f_{e}$; Final image at infinity.
Telescope (image at D): $m = \frac{f _{o}}{f _{e}} (1 + \frac{f _{e}}{D})$; Tube length is then less than f_o + f_e.
Size of image formed by the objective: $h = f_{o} α$; α = angle subtended by the distant object (radians).

How to solve Microscopes and Telescopes questions

Use this method for microscope and telescope numericals.

1Decide the adjustment: normal (final image at infinity) or final image at the near point D = 25 cm.
2Microscope: use the lens formula for the objective to find $v_{o}$ and $m_{o} = \frac{v _{o}}{u _{o}}$ (or use $\frac{L}{f _{o}}$ in normal adjustment).
3Eyepiece: $m_{e} = \frac{D}{f _{e}}$ (normal) or $1 + \frac{D}{f _{e}}$ (image at D). Total m = m_o × m_e.
4For the separation of lenses, add $v_{o}$ to the eyepiece object distance $∣ u_{e} ∣$ found from the lens formula.
5Telescope: $m = \frac{f _{o}}{f _{e}}$ and length $f_{o} + f_{e}$ in normal adjustment; multiply by $(1 + \frac{f _{e}}{D})$ for the near-point case.
6For questions on design, link features to purpose: short f for microscopes, long f and wide aperture for telescope objectives.
7Draw the ray diagram if asked: mark F_o, F_e, the intermediate image and the final image.

Microscopes and Telescopes 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.

Magnifying glass

Suppose a jeweller uses a magnifying glass of focal length 4 cm to look at fine engraving.

(a) Find the magnifying power when the image is formed at the near point (25 cm). (b) Find it when the eye is relaxed (image at infinity). (c) Where should the engraving be placed in case (a)?

Try it first, then show the solutionSolution

(a) $m = 1 + \frac{25}{4} = 7.25$ .
(b) $m = \frac{25}{4} = 6.25$ .
(c) For v = −25 cm: $\frac{1}{u} = \frac{1}{v} - \frac{1}{f} = - \frac{1}{25} - \frac{1}{4} = - \frac{29}{100}$ , so u ≈ −3.45 cm, just inside the focal length.

Answer: (a) 7.25; (b) 6.25; (c) about 3.45 cm from the lens.

Memory hook: A magnifier gives D/f, plus one if you strain to bring the image to 25 cm.

Changing eyepieces on a telescope

Imagine an amateur telescope with an objective of focal length 900 mm, supplied with two eyepieces of focal lengths 25 mm and 10 mm.

(a) Find the magnifying power with each eyepiece in normal adjustment. (b) Find the tube length in each case. (c) Why might the 25 mm eyepiece be preferred for viewing a large, faint object?

Try it first, then show the solutionSolution

(a) $m = \frac{900}{25} = 36$ and $m = \frac{900}{10} = 90$ .
(b) Lengths: 900 + 25 = 925 mm and 900 + 10 = 910 mm.
(c) Lower magnification spreads the gathered light over a smaller image, so a faint extended object looks brighter, and a wider part of the sky is seen.

Answer: (a) 36 and 90; (b) 92.5 cm and 91 cm; (c) it gives a brighter, wider view.

Memory hook: Telescope power = f_objective/f_eyepiece: swap eyepieces to zoom.

Tricks to remember Microscopes and Telescopes

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

Microscope: both short; telescope: long then short
Microscope objective and eyepiece both have short f. Telescope objective long f, eyepiece short f.
Use it when: Design questions.
L over f₀ times D over fₑ
Compound microscope power is two magnifications multiplied.
Use it when: Normal adjustment formula.
Fo over Fe
Telescope power is 'Fo over Fe', and its length is 'Fo plus Fe'.
Use it when: Telescope numericals.
Mirrors don't split colours
Reflection is the same for all colours: no chromatic aberration in a reflecting telescope.
Use it when: Advantages of reflecting telescopes.

Microscopes and Telescopes: solved examples

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

Example 1: Telescope magnification and length · 2 marks · Very short answerBoard level

An astronomical telescope has an objective of focal length 100 cm and an eyepiece of focal length 5 cm. Find its magnifying power and length in normal adjustment.

Try it first, then show the solutionSolution

$m = \frac{f _{o}}{f _{e}} = \frac{100}{5} = 20$ . (1)
$L = f_{o} + f_{e} = 105$ cm. (1)

Answer: 20 and 105 cm.

Example 2: Compound microscope: diagram and magnifying power · 5 marks · Long answerBoard level

(a) Draw a labelled ray diagram of a compound microscope with the final image at the near point. (2)
(b) Derive the expression for its magnifying power, and write it for normal adjustment. (3)

Try it first, then show the solutionSolution

(a) Object AB just beyond F_o; objective forms real, inverted, magnified image A'B' inside the focal length of the eyepiece; eyepiece forms the final virtual image A''B'' at 25 cm; mark F_o, F_e and both lenses. (2)
(b) Magnification by the objective: $m_{o} = \frac{v _{o}}{u _{o}}$ (or $\frac{L}{f _{o}}$ , from similar triangles with the tube length L). (1)
The eyepiece acts as a simple microscope: $m_{e} = 1 + \frac{D}{f _{e}}$ with the image at D. So $m = m_{o} m_{e} = \frac{v _{o}}{u _{o}} (1 + \frac{D}{f _{e}})$ . (1)
Normal adjustment (final image at infinity): $m_{e} = \frac{D}{f _{e}}$ , giving $m \approx \frac{L}{f _{o}} \cdot \frac{D}{f _{e}}$ . (1)

Answer: m = (v_o/u_o)(1 + D/f_e) with the image at D; m ≈ (L/f_o)(D/f_e) in normal adjustment.

Example 3: Compound microscope numerical · 3 marks · Short answerHarder board level

A compound microscope has an objective of focal length 1 cm and an eyepiece of focal length 5 cm. An object is placed 1.1 cm from the objective and the final image is at 25 cm from the eyepiece. Find (a) the magnifying power and (b) the separation of the lenses.

Try it first, then show the solutionSolution

(a) Objective: $\frac{1}{v _{o}} = \frac{1}{1} - \frac{1}{1.1}$ , v_o = 11 cm; $m_{o} = \frac{11}{- 1.1} = - 10$ . (1)
Eyepiece: $m_{e} = 1 + \frac{25}{5} = 6$ ; m = −60 (magnitude 60, final image inverted). (1)
(b) Eyepiece object distance: $\frac{1}{u _{e}} = \frac{1}{- 25} - \frac{1}{5} = - \frac{6}{25}$ , $∣ u_{e} ∣ \approx 4.17$ cm; separation = 11 + 4.17 ≈ 15.2 cm. (1)

Answer: (a) 60 (image inverted); (b) about 15.2 cm.

Example 4: Telescope with image at the near point · 3 marks · Short answerHarder board level

A telescope has f_o = 140 cm and f_e = 5 cm. Find its magnifying power (a) in normal adjustment and (b) when the final image is at 25 cm. (c) Which needs a shorter tube?

Try it first, then show the solutionSolution

(a) $m = \frac{140}{5} = 28$ . (1)
(b) $m = 28 (1 + \frac{5}{25}) = 33.6$ . (1)
(c) Case (b): the eyepiece is moved closer to the intermediate image (about 4.17 cm instead of 5 cm), so the tube is about 144.2 cm instead of 145 cm. (1)

Answer: (a) 28; (b) 33.6; (c) case (b), slightly shorter.

Example 5: Image of the Moon · 5 marks · Long answerHOTS / competency

A telescope objective of focal length 140 cm is used to view the Moon, which subtends an angle of about $9.05 \times 1 0^{- 3}$ rad at the Earth.
(a) Find the diameter of the Moon's image formed by the objective. (2)
(b) With an eyepiece of focal length 5 cm in normal adjustment, find the angle subtended by the final image at the eye. (2)
(c) Why is a large aperture useful even though it does not change the magnifying power? (1)

Try it first, then show the solutionSolution

(a) The image forms at the focal plane: $h = f_{o} α = 140 \times 9.05 \times 1 0^{- 3}$ . (1)
h ≈ 1.27 cm. (1)
(b) $β = m α = \frac{140}{5} \times 9.05 \times 1 0^{- 3}$ . (1)
β ≈ 0.253 rad, about 14.5°. (1)
(c) A larger objective gathers more light, so faint objects become visible and finer detail can be seen. (1)

Answer: (a) about 1.27 cm; (b) about 0.253 rad; (c) more light and finer detail.

Example 6: Reflecting telescope · 4 marks · Case-basedHarder board level

Read the passage and answer.

A reflecting telescope uses a concave mirror of radius of curvature 3 m as its objective, with an eyepiece of focal length 2 cm.
(a) Find the focal length of the mirror and the magnifying power in normal adjustment. (1)
(b) Describe the Cassegrain arrangement. (1)
(c) Give two advantages of a reflecting telescope over a refracting one. (2)

Try it first, then show the solutionSolution

(a) $f_{o} = \frac{R}{2} = 150$ cm; $m = \frac{150}{2} = 75$ . (1)
(b) Light reflected by the large concave primary mirror is intercepted by a small convex secondary mirror, which sends it back through a hole in the primary to the eyepiece behind it. (1)
(c) No chromatic aberration (mirrors reflect all colours alike); spherical aberration can be removed with a paraboloidal mirror; a mirror is lighter and can be supported from behind, so large apertures are practical and cheaper. (2)

Answer: (a) 150 cm and 75; (b) primary concave mirror, convex secondary, image viewed through a hole in the primary; (c) no chromatic aberration, easier to make and support large apertures.

Where students lose marks in Microscopes and Telescopes

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

Mistake: Drawing the intermediate image outside the eyepiece's focal length in a microscope diagram.
Then the eyepiece would form a real image, not a virtual one.
Do this: Place the first image between the eyepiece and its focus (or at its focus for normal adjustment).
Usually costs: 1 mark of the diagram.
Mistake: Using f_o + f_e as the length when the final image is at D.
Applying the normal-adjustment result to every case.
Do this: Find u_e from the lens formula; length = f_o + |u_e|.
Usually costs: 1 mark.
Mistake: Forgetting arrows and labels on ray diagrams.
Rushing.
Do this: Label objective, eyepiece, F_o, F_e, intermediate and final images; put arrows on rays.
Usually costs: ½ to 1 mark.
Mistake: Adding magnifications of the objective and eyepiece.
Confusing with power of lenses in contact.
Do this: Total magnification is the product m_o × m_e.
Usually costs: The answer mark.
Mistake: Listing 'higher magnification' as an advantage of reflecting telescopes.
Magnification depends on focal lengths, not on using a mirror.
Do this: Give: no chromatic aberration, easier to make and support large apertures, reduced spherical aberration.
Usually costs: 1 mark.

Exam tips

Practise the two standard ray diagrams (compound microscope and astronomical telescope) until you can draw them in under three minutes.
Microscopes and telescopes carried 3- and 5-mark questions in recent board papers (for example 2023 55/5/1 Q33, 2025 55/7/1 Q33(a) and 2026 55/5/1 Q33(b)).
Always state which adjustment (normal or near point) your formula is for.
For reflecting telescopes, draw the Cassegrain diagram with the hole in the primary mirror.
Include the sign of m for the microscope only when asked whether the image is erect or inverted.

Practice questions on Microscopes and Telescopes

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 normal adjustment, the magnifying power of an astronomical telescope is
(a) f_o + f_e (b) f_o f_e (c) f_o/f_e (d) f_e/f_o
Hint
Ratio of focal lengths.
Answer
(c) f_o/f_e.
Q2 · 2 marks · Very short answerBoard level
Why does a telescope objective have a large aperture and long focal length, while a microscope objective has a short focal length?
Hint
Think of light gathering and magnification.
Answer
Telescope: a long f_o gives large magnification (f_o/f_e) and a large aperture gathers more light and resolves detail. Microscope: a short f_o makes the objective's magnification L/f_o large for a tiny object placed close to it.
Q3 · 1 mark · Very short answerHarder board level
A compound microscope has f_o = 0.5 cm, f_e = 2.5 cm and tube length 15 cm. Find its magnifying power in normal adjustment.
Hint
(L/f_o)(D/f_e).
Answer
$m = \frac{15}{0.5} \times \frac{25}{2.5} = 300$ .
Q4 · 2 marks · Very short answerHarder board level
An astronomical telescope in normal adjustment has length 92 cm and magnifying power 45. Find the focal lengths of its lenses.
Hint
f_o + f_e = 92, f_o = 45 f_e.
Answer
46f_e = 92, so f_e = 2 cm and f_o = 90 cm.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Large astronomical telescopes are usually reflecting telescopes.
Reason (R): Mirrors have no chromatic aberration and a large mirror can be supported over its entire back.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Compare large lenses and mirrors.
Answer
(a).
Q6 · 2 marks · Very short answerHOTS / competency
The objective and eyepiece of an astronomical telescope are interchanged. What happens to the magnifying power and to the appearance of a distant tree?
Hint
m becomes f_e/f_o.
Answer
The magnifying power becomes f_e/f_o, less than 1, so the tree appears smaller (and farther away) than with the naked eye.
Q7 · 2 marks · Very short answerHOTS / competency
Two magnifying glasses have focal lengths 10 cm and 5 cm. Which gives greater magnification, and by how much, when the image is formed at the near point?
Hint
m = 1 + D/f.
Answer
f = 10 cm: m = 1 + 2.5 = 3.5; f = 5 cm: m = 1 + 5 = 6. The 5 cm lens gives about 1.7 times more magnification.

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.Magnifying power of a simple microscope with image at D?Show answer
1 + D/f.
2.Compound microscope power in normal adjustment?Show answer
(L/f_o)(D/f_e).
3.Telescope power in normal adjustment?Show answer
f_o/f_e.
4.Length of a telescope in normal adjustment?Show answer
f_o + f_e.
5.Why are reflecting telescopes free of chromatic aberration?Show answer
Reflection does not depend on wavelength.
6.Nature of the final image in a compound microscope?Show answer
Virtual, inverted and magnified.

Microscopes and Telescopes: frequently asked questions

What is normal adjustment?

The setting in which the final image is formed at infinity, so the eye views it relaxed. For a telescope, the focus of the objective coincides with the focus of the eyepiece.

Why is the final image in a compound microscope inverted?

The objective forms a real, inverted image; the eyepiece magnifies this without inverting it again, so the final image is inverted relative to the object.

Does a bigger telescope objective increase magnification?

Not by itself: magnification depends on f_o/f_e. A larger aperture gathers more light and shows finer detail, which is why big telescopes are built.

What is the Cassegrain telescope?

A reflecting telescope in which a large concave primary mirror reflects light to a small convex secondary mirror, which sends it back through a hole in the primary to the eyepiece.

Why is a telescope's length about f_o + f_e?

In normal adjustment the image formed by the objective lies at its focal plane, which must also be the focal plane of the eyepiece, so the lenses are f_o + f_e apart.