Light - Reflection & Refraction Topic Hub

Light - Reflection & Refraction — Class 10 Notes

CBSE 2027 — dates, rules and official papers →

Physics · Ch 9Class 10 · CBSE 2026–27High weightage

Light – Reflection and Refraction

What the board actually asks: ray diagrams for mirrors & lenses, the mirror/lens formula numericals (with sign convention), refractive index, power of a lens, and the uses of concave / convex mirrors — the first four carry almost every mark here.

NCERT VERBATIM sourced, word-for-wordAUTHORED LazyTopper teaching scaffolding

Formulae

Mirror formula

\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

Lens formula

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

Magnification

mirror

m = - \frac{v}{u}

· lens

m = \frac{v}{u}

Refractive index

n = \frac{c}{v} = \frac{sin i}{sin r}

Power of lens

P = \frac{1}{f ( m )}

(dioptre, D) · lenses in contact:

P = P_{1} + P_{2} + \dots

Mirror relation

R = 2 f

Definitions the board rewards NCERT VERBATIM

State these in NCERT’s exact words

Laws of reflection of light

NCERT verbatim — memorise this exact wording· Ch 9,

“(i) The angle of incidence is equal to the angle of reflection, and (ii) The incident ray, the normal to the mirror at the point of incidence and the reflected ray, all lie in the same plane.”

What it means: the ray comes in and bounces off at the same angle to the normal (

∠ i = ∠ r

), and the incident ray, normal and reflected ray are all flat in one plane.

Laws of refraction of light (Snell's law)

NCERT verbatim — memorise this exact wording· Ch 9,

“(i) The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane. (ii) The ratio of sine of angle of incidence to the sine of angle of refraction is a constant, for the light of a given colour and for the given pair of media. This law is also known as Snell's law of refraction.”

What it means:

\frac{sin i}{sin r}

stays the same for a given colour of light and a given pair of media — that constant is the refractive index.

Absolute refractive index

NCERT verbatim — memorise this exact wording· Ch 9,

“If medium 1 is vacuum or air, then the refractive index of medium 2 is considered with respect to vacuum. This is called the absolute refractive index of the medium.”

What it means: measured against vacuum/air,

n_{m} = \frac{Speed of light in air}{Speed of light in the medium} = \frac{c}{v}

(NCERT p.149). A larger

n

means light slows more — the medium is optically denser.

Principal focus & focal length of a spherical mirror

NCERT verbatim — memorise this exact wording· Ch 9,

“[Rays parallel to the principal axis] are all meeting/intersecting at a point on the principal axis of the mirror. This point is called the principal focus of the concave mirror. … The distance between the pole and the principal focus of a spherical mirror is called the focal length.”

What it means: parallel rays converge at F (concave) — focal length

f

= pole-to-F distance, and

R = 2 f

Magnification produced by a spherical mirror

NCERT verbatim — memorise this exact wording· Ch 9,

“Magnification produced by a spherical mirror gives the relative extent to which the image of an object is magnified with respect to the object size. It is expressed as the ratio of the height of the image to the height of the object.”

What it means:

m = \frac{h ^{'}}{h} = - \frac{v}{u}

. A negative

m

means a real, inverted image; positive means virtual, erect.

Power of a lens & the dioptre

NCERT verbatim — memorise this exact wording· Ch 9,

“The power of a lens is defined as the reciprocal of its focal length. … The SI unit of power of a lens is 'dioptre'. It is denoted by the letter D. … 1 dioptre is the power of a lens whose focal length is 1 metre. 1D = 1 m⁻¹.”

What it means:

P = \frac{1}{f ( metres )}

. Convex lens → positive power, concave → negative.

Key terms — NCERT verbatim (state word-for-word)

Pole: The centre of the reflecting surface of a spherical mirror is a point called the pole." (Ch 9, )
Centre of curvature: …This point is called the centre of curvature of the spherical mirror." (Ch 9, )
Radius of curvature: The radius of the sphere of which the reflecting surface of a spherical mirror forms a part, is called the radius of curvature of the mirror." (Ch 9, )
Principal axis: [A straight line passing through the pole and the centre of curvature…] This line is called the principal axis." (Ch 9, )
Aperture: The diameter of the reflecting surface of spherical mirror is called its aperture." (Ch 9, )
R = 2f: …the radius of curvature is found to be equal to twice the focal length. We put this as R = 2f." (Ch 9, )
Lens: A transparent material bound by two surfaces, of which one or both surfaces are spherical, forms a lens." (Ch 9, )
Optical centre: The central point of a lens is its optical centre. … A ray of light through the optical centre of a lens passes without suffering any deviation." (p.150–151) (Ch 9, )

Must-know concepts AUTHORED FRAMING

…and where they’re tested

Image formation by spherical mirrors

Trace two standard rays; where they meet is the image. The case below — object beyond C in a concave mirror — gives a real, inverted, diminished image.

NCERT Fig 9.7(b)Concave mirror · object beyond C → real, inverted, diminished image

3–5 mark "draw the ray diagram

Mirror formula & magnification

\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

and

m = \frac{h ^{'}}{h} = - \frac{v}{u}

. Apply the sign convention before substituting.

3-mark numerical

Refraction & refractive index

Light bends toward the normal entering a denser medium.

n = \frac{c}{v} = \frac{sin i}{sin r}

2–3 marks

Refraction through a glass slab & everyday refraction

Light bends when it crosses from one transparent medium into another because its speed changes — towards the normal going air → glass (rarer → denser), away from the normal going glass → air. In a rectangular slab the two bends are equal and opposite, so the emergent ray is parallel to the incident ray but shifted sideways. The same bending is why the bottom of a pond appears raised, a pencil partly in water looks displaced at the surface, and letters under a thick slab seem to rise — explain these qualitatively.

NCERT Fig 9.10Refraction through a glass slab — the ray shifts sideways & emerges parallel (why a pencil in water looks bent)

2-mark "explain why" · everyday-phenomenon; the glass-slab ray diagram (CBSE practical 11)

Lens formula, magnification & power

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

m = \frac{v}{u}

, and power

P = \frac{1}{f ( metres )}

in dioptres (D).

3-mark numerical · power 1-mark

Applications of spherical mirrors and lenses

Concave mirrors give a powerful parallel beam when the source sits at the focus — torches, search-lights and vehicle headlights — and a large erect image when the object is within the focus, which is why they serve as shaving mirrors and dentists' mirrors; large ones concentrate sunlight to produce heat in solar furnaces. Convex mirrors are the rear-view (wing) mirrors of vehicles: the image is always erect (though diminished) and the outward curve gives a much wider field of view than a plane mirror. Lenses: a convex lens as a magnifying glass (watchmakers), spectacle lenses that correct defects of vision (concave for myopia, convex for hypermetropia — Ch 10), and lens combinations in cameras, microscopes and telescopes.

1–2 mark "name the type of mirror used in … and give a reason" (NCERT Exercise Q8: headlights, rear-view mirror, solar furnace)

Worked examples — real solved problems NCERT VERBATIM

Solved exactly as the source does them

Every problem is a real sourced item, quoted verbatim with its page; the step-by-step decoding and mark-logic are LazyTopper-authored coaching.

Example 9.2,

“An object, 4.0 cm in size, is placed at 25.0 cm in front of a concave mirror of focal length 15.0 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Find the nature and the size of the image.”

List with signs (NCERT).

h = + 4.0

cm,

u = - 25.0

cm,

f = - 15.0

cm.Concept: New Cartesian signs

1 mark

Mirror formula.

\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{- 15.0} - \frac{1}{- 25.0} = \frac{- 5.0 + 3.0}{75.0} = \frac{- 2.0}{75.0}

.Concept: substitute with signs

½ mark

Solve.

v = - 37.5

cm → the screen should be placed 37.5 cm in front of the mirror; the image is real.Concept: sign tells nature

½ mark

Size.

h^{'} = - \frac{v h}{u} = - \frac{( - 37.5 ) ( 4.0 )}{( - 25.0 )} = - 6.0

cm → the image is inverted and enlarged.Concept: h′ sign + size

1 mark

Mark-scheme logic (authored): 1 for signed data, 1 for formula + solving

v

, 1 for nature + size with sign. Stating "real, inverted, enlarged" is where the last mark sits.

Example 9.4,

“A 2.0 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. The distance of the object from the lens is 15 cm. Find the nature, position and size of the image. Also find its magnification.”

Problem-solving strategies AUTHORED

Numerical?Write down

u, v, f

with signs before touching the formula. Most marks are lost at the sign step, not the algebra.

Draw the ray diagram"?Use any two of the three standard rays; the image is where the reflected/refracted rays meet.

Mirror vs lens magnification?Mirror

m = - v / u

; lens

m = + v / u

. Don't mix them.

Power question?Convert

f

to metres first, then

P = 1/ f

in dioptres.

Common mistakes that cost marks

Where students lose easy marks

Sign convention errors — the #1 lost mark. Object distance

u

is always negative; concave

f

negative, convex

f

positive.

Mixing mirror and lens magnification — mirror

m = - v / u

, lens

m = + v / u

Forgetting to convert to metres for power —

P = 1/ f

needs

f

in metres, not cm.

Not stating the nature — a numerical answer needs "real/virtual, erect/inverted, magnified/diminished" for full marks.

Formulae

Mirror formula

\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

Lens formula

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

Magnification

mirror

m = - \frac{v}{u}

· lens

m = \frac{v}{u}

Refractive index

n = \frac{c}{v} = \frac{sin i}{sin r}

Power of lens

P = \frac{1}{f ( m )}

(dioptre, D) · lenses in contact:

P = P_{1} + P_{2} + \dots

Mirror relation

R = 2 f

Mindmap

- - Laws of reflection (∠i = ∠r)
  - Concave / convex mirrors
  - Image formation — ray diagrams
  - R = 2f
  - Uses — concave: headlights, shaving / dentist mirror, solar furnace · convex: rear-view mirror
- - 1/v + 1/u = 1/f
  - m = −v/u
  - Sign convention first
- - Bends toward normal (denser)
  - Snell: sin i / sin r = n
  - n = c / v
  - Glass slab: emergent ray parallel, shifted sideways
- - Convex / concave lens
  - 1/v − 1/u = 1/f
  - m = +v/u
  - Power P = 1/f(m), dioptre
  - Uses: magnifying glass, spectacles, camera / microscope / telescope
- - Sign convention errors
  - Mirror vs lens m
  - cm not converted to m (power)
  - Nature not stated

CBSE does not require the mirror/lens formula derivation here — the formal core is the sign convention and ray rules that make every numerical and diagram correct.

NCERT Fig 9.9The New Cartesian Sign Convention for spherical mirrors

Distances (along principal axis)

•All distances measured from the pole / optical centre.
−Distances against the incident light (object side) are negative.
+Distances along the incident light are positive.

What this means in practice

−Object distance $u$ is always negative (real object).
−Concave mirror / concave lens focal length: negative.
+Convex mirror / convex lens focal length: positive.
±Heights above axis +, below axis −.

Ray-diagram construction — the three standard rays (use any two)

3 competency questions from the CBSE board pattern, with the full step-marked solution for each — the marks are shown per step, as the official scheme awards them.

2 marksSection B

Search mirrors are mirrors that are used to look for hidden objects underneath the cars. The hidden objects can be easily spotted as the mirror provides a wider field of view. (a) What type of mirrors are generally used to make search mirrors? (b) With the help of a ray diagram describe the nature of image formed by the type of mirror identified in (a).

Step-marked solution

[1 mark] (a) Convex mirrors — they diverge light and provide a wider field of view than plane or concave mirrors.
[1 mark] (b) Image formed by a convex mirror is always virtual, erect, and diminished. Ray diagram: object in front; one ray parallel to axis appears to diverge from focus behind mirror; another ray towards centre of curvature reflects back along itself. Image forms behind mirror, between pole and focus.

3 marksSection C

Absolute refractive indices of two media P and Q are 1.33 (nP) and 2.52 (nQ) respectively. The speed of light in medium P is 2 × 10^8 m/s. (a) What would be the speed of light in medium Q (vQ)? (b) If the angle of incidence for a ray of light travelling from medium P to Q is 0°, then what will be the path of light in medium Q?

Step-marked solution

[1 mark] nP = c/vP; nQ = c/vQ. Ratio: nP/nQ = vQ/vP ⟹ 1.33/2.52 = vQ / (2 × 10^8).
[1 mark] vQ = (1.33 × 2 × 10^8) / 2.52 = 2.66 / 2.52 × 10^8 ≈ 1.056 × 10^8 m/s.
[1 mark] (b) At normal incidence (angle of incidence = 0°), light enters perpendicular to the interface and travels undeviated (no refraction).

3 marksSection C

A 4 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. The distance of the object from the lens is 15 cm. Find the nature, position and size of the image formed.

Step-marked solution

[0.5 mark] Given: ho = 4 cm, u = -15 cm, f = +20 cm. Using lens formula 1/v - 1/u = 1/f.
[1 mark] 1/v = 1/f + 1/u = 1/20 + 1/(-15) = 3/60 - 4/60 = -1/60 ⟹ v = -60 cm.
[1 mark] Magnification m = v/u = -60 / -15 = +4. So hi = m × ho = 4 × 4 = +16 cm.
[0.5 mark] Nature: image is VIRTUAL, ERECT and MAGNIFIED (formed on the same side as the object, 60 cm from the lens).

Practise Light – Reflection and Refraction competency questions

Official NCERT chapter PDF (jesc109) — ncert.nic.in

Sources: NCERT 2026-27 (tap to view)

Source ledger PROVENANCE

Source edition: NCERT Reprint 2026-27, Class 10 Science, Ch 9. One row per sourced item — every verbatim, example and NCERT figure traces to its page.

Item	Type	Source
Laws of reflection of light	verbatim	NCERT 2026-27 Ch9
Laws of refraction of light (Snell's law)	verbatim	NCERT 2026-27 Ch9
Absolute refractive index	verbatim	NCERT 2026-27 Ch9
Principal focus & focal length of a spherical mirror	verbatim	NCERT 2026-27 Ch9
Magnification produced by a spherical mirror	verbatim	NCERT 2026-27 Ch9
Power of a lens & the dioptre	verbatim	NCERT 2026-27 Ch9
Pole	verbatim	NCERT 2026-27 Ch9
Centre of curvature	verbatim	NCERT 2026-27 Ch9
Radius of curvature	verbatim	NCERT 2026-27 Ch9
Principal axis	verbatim	NCERT 2026-27 Ch9
Aperture	verbatim	NCERT 2026-27 Ch9
R = 2f	verbatim	NCERT 2026-27 Ch9
Lens	verbatim	NCERT 2026-27 Ch9
Optical centre	verbatim	NCERT 2026-27 Ch9
Concave mirror	example	NCERT 2026-27 Ch9 Ex 9.2
Convex lens	example	NCERT 2026-27 Ch9 Ex 9.4
NCERT Fig 9.7(b)	figure	NCERT 2026-27 Ch9 9.7(b)
NCERT Fig 9.10	figure	NCERT 2026-27 Ch9 9.10
NCERT Fig 9.9	figure	NCERT 2026-27 Ch9 9.9

Light – Reflection and Refraction · spec-rendered note · NCERT Reprint 2026-27, Class 10 Science, Ch 9

Practise writing answers the way examiners mark them → Guide