Lenses ICSE Class 10: Ray Diagrams, Lens Formula, Magnification and Power

What Lenses is about

Weightage: CISCE publishes no chapter or unit weightage for ICSE Physics. Lenses is in the 2. Light (iii) part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Physics (Science Paper 1) syllabus, examination year 2027

Key concepts

Types of lenses and technical terms

• Optical centre (O): the point in the lens through which a ray passes undeviated.
• Principal axis: the line through the two centres of curvature.
• Principal focus (F): for a convex lens, the point on the axis where rays parallel to the axis converge after refraction; for a concave lens, the point from which they appear to diverge. A lens has two foci, $F_1$ and $F_2$, one on each side.
• Focal length ($f$): distance from O to F. Focal plane: plane through F perpendicular to the axis.
• A lens acts as a set of prisms: bases towards the centre for a convex lens (converging), towards the edges for a concave lens (diverging).

Worked example

Parallel rays fall on a lens and, after refraction, appear to spread out from a point 20 cm in front of the lens. What kind of lens is it, and what is its focal length?

Try it first, then show the solutionSolution

1. The rays diverge: a concave (diverging) lens.
2. The point they appear to come from is its principal focus, 20 cm away; with the sign convention $f=-20$ cm.

Answer: Concave lens; f = −20 cm

Principal rays for ray diagrams

Use any two of these:
1. A ray parallel to the principal axis: after refraction it passes through $F_2$ (convex) or appears to come from $F_1$ (concave).
2. A ray through (or directed towards) the focus: it emerges parallel to the axis.
3. A ray through the optical centre: goes straight on, undeviated.

The image is where the refracted rays meet (real) or where they appear to meet when extended backwards (virtual). Draw real rays as continuous lines with arrows; draw virtual extensions dashed, without arrows.

Worked example

Which two rays are easiest to draw for an object in front of a convex lens?

Try it first, then show the solutionSolution

1. The ray parallel to the axis, which then passes through $F_2$.
2. The ray through the optical centre, which goes straight through.

Answer: Parallel ray and central ray

Images formed by a convex lens

• Object at infinity: image at $F_2$; real, inverted, highly diminished (point-sized).
• Beyond 2F₁: image between $F_2$ and $2F_2$; real, inverted, diminished (camera).
• At 2F₁: image at $2F_2$; real, inverted, same size.
• Between F₁ and 2F₁: image beyond $2F_2$; real, inverted, magnified (projector).
• At F₁: image at infinity; real, inverted, highly magnified.
• Between F₁ and O: image on the same side as the object; virtual, erect, magnified (magnifying glass).

As the object moves from far away towards $F$, the real image moves away from the lens and grows larger.

Worked example

A convex lens forms a real image 3 times the size of the object. Between which points is the object?

Try it first, then show the solutionSolution

1. A real, magnified image needs the object closer than $2F$ but beyond $F$.
2. So the object lies between $F_1$ and $2F_1$.

Answer: Between F₁ and 2F₁

Images formed by a concave lens

Wherever a real object is placed, a concave lens forms an image that is virtual, erect and diminished, located between the optical centre and $F_1$ on the same side as the object. So its magnification is always positive and less than 1.

A concave lens can never form a real image of a real object, because it always spreads the rays out.

Worked example

Why is a concave lens not used as a reading glass?

Try it first, then show the solutionSolution

1. A reading glass must give an enlarged image.
2. A concave lens always forms a diminished (smaller) virtual image, so it cannot magnify print.

Answer: Its image is always diminished.

Sign convention and the lens formula

New Cartesian sign convention: measure all distances from the optical centre; distances in the direction of the incident light (usually to the right) are positive, those against it are negative; heights above the axis positive, below negative.

• Object distance $u$ is negative for a real object.
• $f$ is positive for a convex lens and negative for a concave lens.

Lens formula: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$.
Magnification: $m=\frac{h^{\prime }}{h}=\frac{v}{u}$. Negative $m$: real, inverted image. Positive $m$: virtual, erect image. $|m|>1$ magnified; $|m|<1$ diminished.

Worked example

An object is 20 cm in front of a convex lens of focal length 15 cm. Find the image position and magnification, and describe the image.

Try it first, then show the solutionSolution

1. $u=-20$ cm, $f=+15$ cm.
2. $\frac{1}{v}=\frac{1}{15}+\frac{1}{-20}=\frac{4-3}{60}=\frac{1}{60}$, so $v=+60$ cm.
3. $m=\frac{60}{-20}=-3$: real, inverted, 3 times magnified, 60 cm behind the lens.

Answer: v = +60 cm; m = −3 (real, inverted, magnified)

Power of a lens

Power measures how strongly a lens converges or diverges light: $P=\frac{1}{f}$, with $f$ in metres. Unit: dioptre (D), $1\text{ D}=1{\text{ m}}^{-1}$.

• Convex lens: positive power. Concave lens: negative power.
• A shorter focal length means a more powerful lens.

Worked example

Find the power of (a) a convex lens of focal length 25 cm, (b) a concave lens of focal length 40 cm.

Try it first, then show the solutionSolution

1. (a) $f=+0.25$ m: $P=\frac{1}{0.25}=+4$ D.
2. (b) $f=-0.40$ m: $P=-2.5$ D.

Answer: (a) +4 D (b) −2.5 D

Magnifying glass and applications

A magnifying glass (simple microscope, reading glass) is a convex lens of short focal length held so that the object is between the lens and its focus. It forms a virtual, erect, magnified image on the same side as the object. Its position and magnification are found from the ray diagram only.

Applications: convex lenses in cameras (object beyond 2F), projectors (object between F and 2F), magnifying glasses, spectacles for long sight; concave lenses in spectacles for short sight and in door peepholes (wide, diminished view).

Worked example

Where must an object be placed relative to a convex lens to use it as a magnifying glass, and what is the nature of the image?

Try it first, then show the solutionSolution

1. Between the optical centre and the principal focus $F_1$.
2. The image is virtual, erect and magnified, on the same side as the object.

Answer: Between O and F₁; virtual, erect, magnified

Physics in real life: Lenses

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.

Topics in this chapter

• Lens Formula and Magnification
• Ray Diagrams for Convex and Concave Lenses
• Power of a Lens (Dioptre)

Lenses formula sheet

Lens formula

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

New Cartesian signs; u negative for a real object.

Magnification

$m=\frac{h^{\prime }}{h}=\frac{v}{u}$

m < 0 real inverted; m > 0 virtual erect.

Power

$P=\frac{1}{f(\text{in m})}$

Dioptre; convex +, concave −.

Signs of f

$f>0\text{convex},f<0\text{concave}$

Fix the sign before substituting.

Same-size real image (convex)

$u=-2f,v=+2f$

Object at 2F gives image at 2F.

Lenses: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Lenses

The question patterns to expect, and the method that works for each.

Tricks to remember Lenses

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

• Convex: far small, near big

  Object far → small image near F; object moves to F → image grows and moves away; inside F → virtual and big.

  Use it when: Image positions for a convex lens.

• Concave = VED

  Concave lens images are always Virtual, Erect, Diminished.

  Use it when: Concave lens questions.

• v minus u

  Lens formula: 1/v − 1/u = 1/f (lens 'minus'; mirror 'plus').

  Use it when: Not mixing up formulas.

• Power in metres

  P = 100 ÷ f (cm) dioptres.

  Use it when: Power numericals.

Lenses: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 10-mark question split into 3 + 3 + 4 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Lenses

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

• ✗Mistake: Interchanging u and v, wrong signs

  CISCE examiners' comments, 2024 and 2025: candidates interchanged the object and image distances, used the wrong sign convention or the wrong formula, and could not identify u and v for a concave lens.

  ✓Do this: Write u (negative for a real object) and v (positive if on a screen, negative if virtual), and the sign of f, before substituting into 1/v − 1/u = 1/f.

  Usually costs: 2 to 3 marks.

• ✗Mistake: Power with f in cm, or as a fraction

  CISCE examiners' comments, 2024: candidates left the answer as a fraction and did not convert the focal length to metres before finding power.

  ✓Do this: P = 1/f with f in metres (or P = 100/f with f in cm); give decimals.

  Usually costs: 1 mark.

• ✗Mistake: Careless ray diagrams

  CISCE examiners' comments, 2025: no arrows on real rays, arrows on virtual rays, different scales on the two sides of the lens, the object not at 2F, and dotted lines for real rays.

  ✓Do this: Equal F and 2F spacing on both sides; real rays continuous with arrows; virtual rays dashed without arrows.

  Usually costs: 1 to 2 marks.

• ✗Mistake: Not reading the condition in the question

  CISCE examiners' comments, 2025: asked what happens when ONLY the lens is replaced by one of greater focal length, candidates changed the object position instead.

  ✓Do this: Keep everything else fixed; a longer f brings the object relatively closer to F, so the real image grows.

  Usually costs: 1 mark.

• ✗Mistake: Magnification inequalities

  CISCE examiners' comments, 2024 and 2025: candidates confused m < 1 with m > 1 for a concave lens, took magnification as O : I instead of I : O, and could not link a size ratio to the object position.

  ✓Do this: m = image height ÷ object height. Concave lens: m < 1. Convex real image smaller than object: object beyond 2F.

  Usually costs: 1 mark.

• ✗Mistake: Vague answers about reading glasses

  CISCE examiners' comments, 2024: candidates could not give the exact object position for a reading glass and explained the concave lens by its diverging nature instead of the image size.

  ✓Do this: Object between O and F; a concave lens always gives a diminished image.

  Usually costs: 1 mark each.

Practice questions on Lenses

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

1. Q1 · 1 mark · MCQBoard level

   The SI unit of power of a lens is:
   (a) watt
   (b) dioptre
   (c) metre
   (d) joule

   Hint

   P = 1/f.

   Answer

   (b)

2. Q2 · 1 mark · MCQBoard level

   An object at $2F_1$ of a convex lens forms an image that is:
   (a) at $F_2$, diminished
   (b) at $2F_2$, same size, inverted
   (c) beyond $2F_2$, magnified
   (d) virtual and erect

   Hint

   Symmetric case.

   Answer

   (b)

3. Q3 · 1 mark · MCQHarder board level

   A convex lens forms a real image whose height is one-third that of the object. The object is:
   (a) between O and F
   (b) at F
   (c) between F and 2F
   (d) beyond 2F

   Hint

   Real and diminished.

   Answer

   (d)

4. Q4 · 1 mark · Assertion-reasonBoard level

   Assertion (A): A convex lens can form a virtual image.
   Reason (R): When the object is between the optical centre and the focus, the refracted rays diverge and appear to meet on the object's side.
   (a) (A) is true and (R) is false.
   (b) (A) is false and (R) is true.
   (c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
   (d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).

   Hint

   Magnifying glass.

   Answer

   (c)

5. Q5 · 2 marks · Short answerHarder board level

   A lens has a power of +1.25 D. Find its focal length and state its type.

   Hint

   f = 1/P.

   Answer

   $f=\frac{1}{1.25}=0.8$ m $=+80$ cm; convex (converging) lens.

6. Q6 · 2 marks · Short answerHarder board level

   Explain why a concave lens can never form a real image of a real object.

   Hint

   What do the rays do after the lens?

   Answer

   A concave lens always makes rays from a real object diverge; diverging rays never meet, so they only appear to come from a point in front of the lens: the image is always virtual.

7. Q7 · 3 marks · Long answerBoard level

   Define (a) principal focus of a convex lens, (b) principal focus of a concave lens, (c) optical centre.

   Hint

   Converge, appear to diverge, undeviated.

   Answer

   (a) The point on the axis where rays parallel to the axis converge after refraction. (b) The point on the axis from which rays parallel to the axis appear to diverge after refraction. (c) The point in the lens through which a ray passes without deviation.

8. Q8 · 3 marks · Long answerHarder board level

   An object is placed 10 cm from a convex lens of focal length 15 cm. Find the position and magnification of the image and state its nature.

   Hint

   Object inside F.

   Answer

   $\frac{1}{v}=\frac{1}{15}-\frac{1}{10}=-\frac{1}{30}$, $v=-30$ cm; $m=\frac{-30}{-10}=+3$: virtual, erect, magnified, 30 cm on the object's side.

9. Q9 · 3 marks · Long answerHOTS / competency

   An object 30 cm from a lens gives an erect image half its size, 15 cm from the lens on the same side. (a) Is the lens convex or concave? (b) Find its focal length and power.

   Hint

   Erect and diminished.

   Answer

   (a) Concave (erect and diminished). (b) $\frac{1}{f}=\frac{1}{-15}-\frac{1}{-30}=-\frac{1}{30}$, $f=-30$ cm; $P\approx -3.33$ D.

10. Q10 · 3 marks · Long answerHarder board level

    An object is moved from far away towards the focus of a convex lens. Describe how the position and size of its real image change.

    Hint

    Far → F, near → infinity.

    Answer

    The image moves from near F₂ through 2F₂ to beyond 2F₂ and finally to infinity; it grows from point-sized to diminished, same size at 2F, magnified, and highly magnified.

11. Q11 · 1 mark · MCQHOTS / competency

    A convex lens of focal length 20 cm forms a real image of the same size as the object when the object is placed at:
    (a) 10 cm
    (b) 20 cm
    (c) 40 cm
    (d) 80 cm

    Hint

    u = 2f.

    Answer

    (c)

12. Q12 · 4 marks · Long answerHOTS / competency

    A convex lens forms a real image three times the size of the object on a screen. The distance between the object and the screen is 80 cm. Find the object and image distances and the focal length of the lens.

    Hint

    v = 3|u| and |u| + v = 80.

    Answer

    $|u|+3|u|=80$, so $|u|=20$ cm and $v=60$ cm ($u=-20$ cm). $\frac{1}{f}=\frac{1}{60}+\frac{1}{20}=\frac{4}{60}$, $f=15$ 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. 1.Lens formula?Show answer

   1/v − 1/u = 1/f

2. 2.Magnification of a lens?Show answer

   m = v/u = h′/h

3. 3.Sign of f for a concave lens?Show answer

   Negative.

4. 4.Unit of power?Show answer

   Dioptre (D).

5. 5.Image when the object is at 2F of a convex lens?Show answer

   At 2F, real, inverted, same size.

6. 6.Object position for a magnifying glass?Show answer

   Between the optical centre and F.

7. 7.Nature of a concave lens image?Show answer

   Virtual, erect, diminished.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you draw all six convex-lens cases?
• ☐Do you mark F and 2F at equal spacing?
• ☐Do you assign signs before substituting?
• ☐Can you find f from u and v for both lenses?
• ☐Do you convert f to metres for power?
• ☐Can you predict the image change when f changes?

Lenses: frequently asked questions