PhysicsICSE Class 10 · Chapter 7

Refraction of Light ICSE Class 10: Laws, Refractive Index, Glass Block and Prism

Updated 14 October 2026 · Fact-checked

Refraction is the bending of light when it passes from one medium to another because its speed changes. In ICSE Class 10 you use Snell's law sin i/sin r = μ, refractive index μ = c/v, real and apparent depth, lateral displacement in a glass block and deviation in a prism, with ray diagrams.

What Refraction of Light is about

Light travels at different speeds in different media: fastest in vacuum (about $3 \times 1 0^{8}$ m/s), slower in water and glass. When it crosses a boundary obliquely, this change of speed makes it bend: this is refraction. Refraction explains why a pool looks shallower than it is, why a stick in water looks bent, and how a prism splits light.

The CISCE 2027 syllabus covers: refraction through a glass block and a triangular prism; partial reflection and refraction; the laws of refraction; the effect on speed, wavelength and frequency; conditions for a ray to pass undeviated; refractive index $μ = \frac{c}{V}$ and $V = f λ$ , with values for water, glass and diamond; lateral displacement; multiple images in a thick mirror; deviation through a prism; real and apparent depth and the apparent bending of a stick. Simple numerical problems and approximate ray diagrams are required.

The CISCE examiners' comments for 2024 and 2025 list: ray diagrams without arrows or with dotted lines, wrong factors for deviation, thinking μ depends on the angle of incidence, not naming the principle of reversibility, and missing units for depth.

Weightage: CISCE publishes no chapter or unit weightage for ICSE Physics. Refraction of Light is in the 2. Light (i) 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

Refraction, its cause and partial reflection

Refraction is the change in direction of light when it passes obliquely from one transparent medium to another. The cause is the change in the speed of light.

Rarer to denser (air → glass): the ray bends towards the normal ( $r < i$ ).
Denser to rarer (glass → air): the ray bends away from the normal ( $r > i$ ).

At every boundary some light is also reflected (partial reflection), so we can see faint reflections in a window.

A ray passes undeviated if it falls normally on the boundary ( $i = 0$ , so $r = 0$ ), or if the two media have the same refractive index.

Worked example

A ray goes from glass into air at an angle of incidence of 30°. Will it bend towards or away from the normal? What happens if it falls along the normal?

Try it first, then show the solutionSolution

Glass is denser than air: the ray speeds up and bends away from the normal, so $r > 3 0^{\circ}$ .
Along the normal ( $i = 0$ ) it goes straight through, undeviated.

Answer: Away from the normal; undeviated at normal incidence

Laws of refraction and refractive index

1. The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.
2. Snell's law: for a given pair of media and a given colour, $\frac{s i n i}{s i n r}$ is constant. This constant is the refractive index of the second medium with respect to the first, $_{1} μ_{2}$ .

Absolute refractive index: $μ = \frac{c}{V}$ (speed in vacuum ÷ speed in the medium). It has no unit and is greater than 1.
Typical values: water 1.33, glass 1.5, diamond 2.42. Speeds: water $\approx 2.25 \times 1 0^{8}$ m/s, glass $2 \times 1 0^{8}$ m/s, diamond $\approx 1.24 \times 1 0^{8}$ m/s.

$μ$ depends on the medium, the wavelength (colour) of light (more for violet, less for red) and the temperature; it does not depend on the angle of incidence.

Worked example

A ray in air strikes glass of refractive index $3$ at an angle of incidence of $6 0^{\circ}$ . Find the angle of refraction.

Try it first, then show the solutionSolution

$sin r = \frac{s i n 6 0 ^{\circ}}{3} = \frac{3 /2}{3} = \frac{1}{2}$ .
$r = 3 0^{\circ}$ .

Answer: 30°

Speed, wavelength and frequency on refraction

When light enters a new medium:

**frequency $f$ does not change** (it is fixed by the source);
speed changes: $V = \frac{c}{μ}$ ;
wavelength changes in the same ratio, since $V = f λ$ : $λ_{medium} = \frac{λ _{air}}{μ}$ .

In a denser medium both speed and wavelength decrease.

Worked example

Light of wavelength 600 nm in air enters glass of refractive index 1.5. Find its speed, wavelength and frequency in glass.

Try it first, then show the solutionSolution

Speed $= \frac{3 \times 1 0 ^{8}}{1.5} = 2 \times 1 0^{8}$ m/s.
Wavelength $= \frac{600}{1.5} = 400$ nm.
Frequency unchanged: $\frac{3 \times 1 0 ^{8}}{600 \times 1 0 ^{- 9}} = 5 \times 1 0^{14}$ Hz.

Answer: 2 × 10⁸ m/s; 400 nm; 5 × 10¹⁴ Hz

Relative refractive index and reversibility

Refractive index of medium b with respect to medium a: $_{a} μ_{b} = \frac{μ _{b}}{μ _{a}} = \frac{V _{a}}{V _{b}}$ .

Principle of reversibility: if a ray's direction is reversed, it retraces its path. So $_{a} μ_{b} \times_{b} μ_{a} = 1$ , i.e. $_{b} μ_{a} = \frac{1}{_{a} μ _{b}}$ .

Worked example

The refractive indices of a glass and of water are 1.6 and $\frac{4}{3}$ . Find the refractive index of glass with respect to water, and of water with respect to glass. Name the principle used for the second.

Try it first, then show the solutionSolution

$_{w} μ_{g} = \frac{1.6}{4/3} = 1.2$ .
$_{g} μ_{w} = \frac{1}{1.2} \approx 0.83$ .
Principle of reversibility of light.

Answer: 1.2; about 0.83; principle of reversibility

Glass block: emergent ray and lateral displacement

A ray passing through a parallel-sided glass block is refracted twice. Because the faces are parallel, the emergent ray is parallel to the incident ray but shifted sideways. This perpendicular shift is the lateral displacement.

Lateral displacement increases with: the thickness of the block, the angle of incidence, and the refractive index (so it is greater for violet than for red).

A thick glass mirror gives multiple images: light is partly reflected at the front surface and partly at the silvered back, with further reflections inside. The second image (from the silvered surface) is the brightest.

Worked example

State two factors on which the lateral displacement of a ray through a glass block depends, and the relation between the emergent and incident rays.

Try it first, then show the solutionSolution

Thickness of the block; angle of incidence (also the refractive index / colour).
The emergent ray is parallel to the incident ray.

Answer: Thickness and angle of incidence; emergent ∥ incident

Real and apparent depth

An object under water (or inside glass) seen from above appears raised, because rays from it bend away from the normal as they leave the denser medium. For viewing nearly vertically:
$μ = \frac{real depth}{apparent depth}$ .
Shift (apparent rise) $=$ real depth $-$ apparent depth $=$ real depth $(1 - \frac{1}{μ})$ .

The same effect makes a stick dipped in water look bent at the surface and shorter under water. Since $μ$ is larger for blue than for red, an object looks shallower in blue light.

Worked example

A coin lies at the bottom of a beaker filled with water to a depth of 24 cm ( $μ = \frac{4}{3}$ ). Find its apparent depth and the apparent rise.

Try it first, then show the solutionSolution

Apparent depth $= \frac{24}{4/3} = 18$ cm.
Rise $= 24 - 18 = 6$ cm.

Answer: 18 cm; 6 cm

Refraction through a prism

A triangular prism has two refracting faces inclined at the angle of the prism $A$ . A ray bends towards the base at each face, so the emergent ray is deviated from the incident direction by the angle of deviation $δ$ .

$r_{1} + r_{2} = A$ and $δ = i + e - A$ .
$δ$ depends on: the angle of incidence, the angle of the prism, the material (μ) and the colour of light (violet is deviated most, red least).
At minimum deviation, $i = e$ , $r_{1} = r_{2} = \frac{A}{2}$ , and in an equilateral prism the ray inside is parallel to the base: $δ_{m} = 2 i - A$ .

Worked example

A ray falls at $5 0^{\circ}$ on an equilateral prism and passes at minimum deviation. Find $r_{1}$ , the angle of emergence, $δ_{m}$ and the refractive index. ( $sin 5 0^{\circ} = 0.766$ )

Try it first, then show the solutionSolution

$r_{1} = r_{2} = 3 0^{\circ}$ ; $e = i = 5 0^{\circ}$ .
$δ_{m} = 2 \times 5 0^{\circ} - 6 0^{\circ} = 4 0^{\circ}$ .
$μ = \frac{s i n 5 0 ^{\circ}}{s i n 3 0 ^{\circ}} = \frac{0.766}{0.5} \approx 1.53$ .

Answer: 30°; 50°; 40°; about 1.53

Physics in real life: Refraction of Light

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 swimming pool looks shallower

Suppose a swimming pool is 2.4 m deep and filled with water of refractive index $\frac{4}{3}$ .

How deep does it appear when viewed from directly above?

Try it first, then show the solutionSolution

Apparent depth $= \frac{2.4}{4/3} = 1.8$ m.
It looks 0.6 m shallower than it really is.

Answer: 1.8 m

Memory hook: Water lies about depth: divide by μ.

Speed of light in diamond

Imagine light entering a diamond of refractive index 2.4.

Find the speed of light inside the diamond.

Try it first, then show the solutionSolution

$V = \frac{c}{μ} = \frac{3 \times 1 0 ^{8}}{2.4}$ .
$V = 1.25 \times 1 0^{8}$ m/s.

Answer: 1.25 × 10⁸ m/s

Memory hook: Bigger μ, slower light.

A bent straw

Suppose you put a straw in a glass of water and look at it from the side, slightly above the surface.

Why does the straw look bent at the water surface?

Try it first, then show the solutionSolution

Light from the submerged part bends away from the normal as it leaves the water.
Each point under water appears raised, so the submerged part looks higher and the straw appears bent at the surface.

Answer: Refraction raises the image of the submerged part.

Memory hook: Under water things float up in the image.

Topics in this chapter

Refraction of Light formula sheet

Snell's law: $\frac{s i n i}{s i n r} =_{1} μ_{2}$; For a given pair of media and colour.
Absolute refractive index: $μ = \frac{c}{V}$; c = 3 × 10⁸ m/s.
Wave equation: $V = f λ$; f unchanged on refraction; λ ∝ V.
Relative index and reversibility: $_{a} μ_{b} = \frac{μ _{b}}{μ _{a}},_{a} μ_{b} \times_{b} μ_{a} = 1$; Principle of reversibility.
Real and apparent depth: $μ = \frac{real depth}{apparent depth}$; Shift = real depth (1 − 1/μ).
Prism: $r_{1} + r_{2} = A, δ = i + e - A$; Minimum deviation: i = e, δ_m = 2i − A.

Refraction of Light: diagrams to practise

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

Refraction through a rectangular glass block (lateral displacement)

Draw the path of a ray through a glass block and mark the lateral displacement.

Draw

A rectangle for the block.
An incident ray meeting the top face obliquely, with a dashed normal.
The refracted ray inside, bending towards the normal.
The emergent ray from the bottom face, bending away from the normal and parallel to the incident ray.
The incident ray extended (dashed) through the block; the perpendicular distance between it and the emergent ray.

Label

Incident ray, refracted ray, emergent ray (with arrows)
Normals at both faces
Angles i, r, r′, e (with i = e)
Lateral displacement
Air and glass

Watch out: Use continuous lines with arrows for rays; dashed lines only for normals and extensions.

Real and apparent depth

Explain why a coin or a pool looks shallower.

Draw

A water surface (horizontal line) and the bottom.
The real object O at the bottom.
Two rays from O meeting the surface: one along the normal (straight up), one slightly oblique bending away from the normal into air.
The oblique refracted ray extended backwards (dashed) to meet the normal ray at I.
An eye looking down from above.

Label

Real object O and image I
Real depth and apparent depth
Normal
Water (denser) and air (rarer)
Eye

Watch out: The image is above the object: apparent depth is less than real depth.

Ray through an equilateral prism at minimum deviation

Complete the path of a ray; mark the angles.

Draw

An equilateral triangle with the base horizontal.
An incident ray on one face, with a dashed normal.
The refracted ray inside, parallel to the base.
The emergent ray from the second face, bending away from the normal (towards the base).
The incident and emergent rays extended (dashed) to meet, showing δ.

Label

Angle of prism A
i, r₁, r₂, e (with i = e, r₁ = r₂)
Angle of deviation δ
Normals at both faces
Arrows on all rays

Watch out: Both refractions bend the ray towards the base. Do not show total internal reflection unless the angle needs it.

How the board asks Refraction of Light

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

Question type	How it is asked	Method
MCQ and fill-in-the-blanks	What does not change on refraction; what μ depends on; direction of bending.	Frequency is unchanged; μ depends on medium, colour, temperature, not on i.
Short numericals (2-3 marks)	Speed from μ; relative μ; real and apparent depth; Snell's law.	μ = c/V; ₐμ_b = μ_b/μ_a; μ = real ÷ apparent; give units.
Glass block and lateral displacement	Draw the ray path; factors affecting lateral displacement; which ray is displaced more.	Emergent ∥ incident; more bending or thicker block gives more displacement.
Prism ray diagrams (3-4 marks)	Complete the path through a prism; factors affecting deviation; minimum deviation.	Bend towards the base at each face; at minimum deviation the inside ray is parallel to the base.
Applications and reasons	Bent stick, raised coin, multiple images in a thick mirror, colour effects.	Bending away from the normal on leaving the denser medium; partial reflections.

Tricks to remember Refraction of Light

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

FAST: Frequency Always STays
On refraction, frequency stays the same; speed and wavelength change.
Use it when: What changes on refraction.
Slow bends to the normal
Light slowing down (into a denser medium) bends towards the normal.
Use it when: Direction of bending.
Real over apparent
μ = real ÷ apparent: the bigger number goes on top.
Use it when: Depth numericals.
Violet is bent most
VIBGYOR: violet has the highest μ, bends and deviates most; red least.
Use it when: Colour effects in blocks and prisms.

Refraction of Light: board-level solved questions

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

Example 1: Leaving glass · 1 mark · MCQBoard level

When light passes from glass into air:
(a) its speed decreases
(b) its speed increases
(c) its frequency increases
(d) its wavelength decreases

Try it first, then show the solutionSolution

Air is optically rarer: speed (and wavelength) increase; frequency is unchanged.

Answer: (b)

Example 2: Normal incidence · 1 mark · Assertion-reasonBoard level

Assertion (A): A ray of light falling normally on a glass surface passes into the glass without bending.
Reason (R): For $i = 0^{\circ}$ , Snell's law gives $r = 0^{\circ}$ .
(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).

Try it first, then show the solutionSolution

$sin 0 = 0$ , so $sin r = 0$ and $r = 0$ .

Answer: (c)

Example 3: Speed in water · 2 marks · Short answerBoard level

The refractive index of water is $\frac{4}{3}$ . Find the speed of light in water. ( $c = 3 \times 1 0^{8}$ m/s)

Try it first, then show the solutionSolution

$V = \frac{c}{μ} = \frac{3 \times 1 0 ^{8}}{4/3}$
$= 2.25 \times 1 0^{8}$ m/s

Answer: 2.25 × 10⁸ m/s

Example 4: Snell's law · 3 marks · Short answerHarder board level

A ray in air is incident at $6 0^{\circ}$ on a glass surface. The refractive index of the glass is $3$ . Find (a) the angle of refraction, (b) the angle through which the ray turns at the surface.

Try it first, then show the solutionSolution

(a) $sin r = \frac{s i n 6 0 ^{\circ}}{3} = \frac{1}{2}$ , so $r = 3 0^{\circ}$ .
(b) Deviation at the surface $= i - r = 3 0^{\circ}$ .
The ray bends towards the normal.

Answer: (a) 30° (b) 30°

Example 5: Relative refractive index and speed · 3 marks · Short answerHarder board level

The absolute refractive indices of diamond and of a glass are 2.4 and 1.5. (a) Find the refractive index of diamond with respect to the glass. (b) Light travels at $2 \times 1 0^{8}$ m/s in the glass; find its speed in diamond. (c) Find the refractive index of the glass with respect to diamond, and name the principle you use.

Try it first, then show the solutionSolution

(a) $_{g} μ_{d} = \frac{μ _{d}}{μ _{g}} = \frac{2.4}{1.5} = 1.6$ .
(b) $_{g} μ_{d} = \frac{V _{g}}{V _{d}}$ , so $V_{d} = \frac{2 \times 1 0 ^{8}}{1.6} = 1.25 \times 1 0^{8}$ m/s.
(c) $_{d} μ_{g} = \frac{1}{1.6} = 0.625$ , by the principle of reversibility of light.

Answer: (a) 1.6 (b) 1.25 × 10⁸ m/s (c) 0.625; principle of reversibility

Example 6: A fish seen by a kingfisher · 2 marks · Long answerBoard level

A fish swims 1.2 m below the surface of a still pond. A kingfisher looks at it from directly above. If $μ = \frac{4}{3}$ for water, how deep does the fish appear to be, and by how much does it appear raised?

Try it first, then show the solutionSolution

Apparent depth $= \frac{real depth}{μ} = \frac{1.2}{4/3} = 0.9$ m.
Apparent rise $= 1.2 - 0.9 = 0.3$ m, so the bird must dive deeper than the fish appears to be.

Answer: 0.9 m; raised by 0.3 m

Example 7: Wavelength in water · 3 marks · Long answerHarder board level

Blue light of wavelength 450 nm in air enters water of refractive index $\frac{4}{3}$ . Find (a) its frequency, (b) its speed in water, (c) its wavelength in water. ( $c = 3 \times 1 0^{8}$ m/s)

Try it first, then show the solutionSolution

(a) $f = \frac{c}{λ} = \frac{3 \times 1 0 ^{8}}{450 \times 1 0 ^{- 9}} \approx 6.67 \times 1 0^{14}$ Hz, the same in air and in water.
(b) $V = \frac{c}{μ} = \frac{3 \times 1 0 ^{8}}{4/3} = 2.25 \times 1 0^{8}$ m/s.
(c) $λ_{w} = \frac{λ}{μ} = \frac{450}{4/3} = 337.5$ nm.

Answer: (a) about 6.67 × 10¹⁴ Hz (b) 2.25 × 10⁸ m/s (c) 337.5 nm

Example 8: Prism at minimum deviation · 4 marks · Long answerHOTS / competency

A ray of monochromatic light falls on one face of an equilateral glass prism at an angle of incidence of $5 0^{\circ}$ and passes through at minimum deviation. (a) Find the angles of refraction at the two faces and the angle of emergence. (b) Find the angle of minimum deviation. (c) Find the refractive index of the glass. (d) How would the deviation change if violet light replaced red light at the same angle of incidence? ( $sin 5 0^{\circ} = 0.766$ )

Try it first, then show the solutionSolution

(a) $r_{1} = r_{2} = \frac{6 0 ^{\circ}}{2} = 3 0^{\circ}$ ; $e = 5 0^{\circ}$ .
(b) $δ_{m} = 5 0^{\circ} + 5 0^{\circ} - 6 0^{\circ} = 4 0^{\circ}$ .
(c) $μ = \frac{0.766}{0.5} \approx 1.53$ .
(d) Glass has a larger μ for violet, so violet light is deviated more.

Answer: (a) 30°, 30°, 50° (b) 40° (c) about 1.53 (d) greater for violet

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.

Safety at the swimming pool

Suppose a pool is 2.4 m deep at the deep end, and the water has refractive index $\frac{4}{3}$ .

(i) · 3 marks
Find the apparent depth of the deep end seen from directly above.
Answer
Answer: 1.8 m
$\frac{2.4}{4/3} = 1.8$ m.
(ii) · 3 marks
Explain, with the idea of refraction, why a swimmer standing at the edge may misjudge the depth.
Answer
Answer: Rays bend away from the normal on leaving water
Light from the bottom bends away from the normal as it passes from water to air; traced back, the rays seem to come from a point higher up. The pool appears 0.6 m shallower than it is, which can be dangerous for divers.
(iii) · 4 marks
Would the pool appear deeper or shallower in blue light than in red light? Explain.
Answer
Answer: Shallower in blue light
Water has a larger refractive index for blue than for red. Apparent depth = real depth ÷ μ, so a larger μ gives a smaller apparent depth: the pool looks shallower in blue light.

Measuring the refractive index of a glass block

In a laboratory, a student traces a ray through a rectangular glass block. She measures the angle of incidence as $4 5^{\circ}$ and the angle of refraction as $2 8^{\circ}$ . ( $sin 4 5^{\circ} = 0.7071$ , $sin 2 8^{\circ} = 0.4695$ )

(i) · 3 marks
Find the refractive index of the glass (to 2 decimal places).
Answer
Answer: About 1.51
$μ = \frac{0.7071}{0.4695} \approx 1.51$ .
(ii) · 3 marks
Find the speed of light in the glass.
Answer
Answer: About 1.99 × 10⁸ m/s
$V = \frac{3 \times 1 0 ^{8}}{1.506} \approx 1.99 \times 1 0^{8}$ m/s.
(iii) · 4 marks
State how the emergent ray is related to the incident ray, and two ways the student could increase the lateral displacement.
Answer
Answer: Parallel; thicker block, larger angle of incidence
The emergent ray is parallel to the incident ray because the two faces are parallel. Lateral displacement increases with a thicker block and with a larger angle of incidence (also with a material of higher μ).

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.

Light travels 30 cm in vacuum in the same time as it travels 20 cm in a medium. Find the refractive index of the medium. (2 marks)

Step	Marks
In equal times, distance ∝ speed: $\frac{c}{V} = \frac{30}{20}$	1
$μ = 1.5$	1

An observer looking vertically down into a tank sees its bottom 30 cm below the surface. If $μ = \frac{4}{3}$ , find the real depth, and how much water must be added for the bottom to appear 45 cm deep. (3 marks)

Step	Marks
Real depth $= 30 \times \frac{4}{3} = 40$ cm	1
New real depth $= 45 \times \frac{4}{3} = 60$ cm	1
Add $60 - 40 = 20$ cm of water	1

A ray passes through an equilateral prism at minimum deviation with an angle of incidence of $5 0^{\circ}$ . Draw the ray path and find $r_{1}$ and the angle of minimum deviation. (4 marks)

Step	Marks
Diagram: ray inside parallel to the base, arrows, normals	1
Angles $i$ , $e$ and $δ$ marked, with $i = e$	1
$r_{1} = \frac{A}{2} = 3 0^{\circ}$	1
$δ_{m} = 2 i - A = 4 0^{\circ}$	1

Where students lose marks in Refraction of Light

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

Mistake: Ray diagrams without arrows or with dotted rays
CISCE examiners' comments, 2024: in completing the path through a prism at minimum deviation, candidates missed arrows, did not draw the ray inside parallel to the base or mark e equal to i, used dotted lines for rays, or showed total internal reflection.
Do this: Rays: continuous lines with arrows. Normals: dashed. At minimum deviation in an equilateral prism, the inside ray is parallel to the base and e = i.
Usually costs: 1 to 2 marks.
Mistake: Wrong factors for deviation
CISCE examiners' comments, 2024: candidates gave more than two factors, confused deviation with lateral displacement (writing 'thickness'), or wrote 'colour of prism' or 'refractive index of light'.
Do this: Deviation depends on the angle of incidence, the angle of the prism, the material (μ) and the colour (wavelength) of light.
Usually costs: 1 mark each.
Mistake: Thinking μ depends on the angle of incidence
CISCE examiners' comments, 2025: because μ = sin i/sin r, candidates thought μ depends on the angle of incidence; others guessed which quantity stays the same on refraction.
Do this: Changing i changes r but not μ. μ depends on the medium, the wavelength and the temperature; frequency never changes on refraction.
Usually costs: 1 mark.
Mistake: Not naming the principle of reversibility
CISCE examiners' comments, 2024: candidates wrote 'law of refraction', 'law of reflection' or 'Snell's law' instead of the principle of reversibility when finding the refractive index of water with respect to glass from that of glass with respect to water.
Do this: aμb × bμa = 1 comes from the principle of reversibility. 'Name' needs only the name.
Usually costs: 1 mark.
Mistake: Speed reasoning and units
CISCE examiners' comments, 2024 and 2025: candidates did not realise light covers the greatest distance in vacuum in a given time, made errors finding μ from speeds, and gave the real depth of a bubble without a unit.
Do this: Light is fastest in vacuum; μ = c/V; always write cm or m with depths.
Usually costs: 1 mark each.
Mistake: Not linking bending, speed and colour
CISCE examiners' comments, 2025: comparing two rays through a block, candidates did not see that the ray bent more travels more slowly, has a shorter wavelength, is displaced more and scatters more.
Do this: More bending ⇒ larger μ ⇒ lower speed ⇒ shorter wavelength (towards violet).
Usually costs: Up to 3 marks.

Practice questions on Refraction of Light

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
The refractive index of ordinary glass is about:
(a) 1.0
(b) 1.33
(c) 1.5
(d) 2.42
Hint
Water 1.33, diamond 2.42.
Answer
(c)
Q2 · 1 mark · MCQBoard level
A ray of light passing obliquely from air into water:
(a) bends towards the normal
(b) bends away from the normal
(c) does not bend
(d) is totally reflected
Hint
Rarer to denser.
Answer
(a)
Q3 · 1 mark · MCQHarder board level
The refractive index of a medium does NOT depend on:
(a) the nature of the medium
(b) the wavelength of light
(c) the temperature
(d) the angle of incidence
Hint
Changing i only changes r.
Answer
(d)
Q4 · 1 mark · Assertion-reasonBoard level
Assertion (A): A pencil partly dipped in water appears bent at the surface.
Reason (R): Light from the immersed part bends away from the normal on leaving the water, so that part appears raised.
(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
Apparent depth.
Answer
(c)
Q5 · 2 marks · Short answerHarder board level
The refractive index of diamond is 2.4. Find the speed of light in diamond.
Hint
V = c/μ.
Answer
$\frac{3 \times 1 0 ^{8}}{2.4} = 1.25 \times 1 0^{8}$ m/s
Q6 · 2 marks · Short answerHarder board level
In the same time, light travels 30 cm in vacuum and 20 cm in a transparent medium. Find the refractive index of the medium and the speed of light in it.
Hint
Distance ∝ speed in equal times.
Answer
$μ = \frac{30}{20} = 1.5$ ; $V = 2 \times 1 0^{8}$ m/s.
Q7 · 3 marks · Long answerBoard level
(a) State the two laws of refraction. (b) Define the absolute refractive index of a medium.
Hint
Same plane; Snell's law.
Answer
(a) The incident ray, refracted ray and normal lie in one plane; sin i/sin r is constant for a given pair of media and colour. (b) The ratio of the speed of light in vacuum to its speed in the medium, μ = c/V.
Q8 · 3 marks · Long answerHarder board level
A coin at the bottom of a beaker of water 16 cm deep appears raised by 4 cm when seen from above. (a) Find the refractive index of water. (b) If the water is replaced by the same depth of a liquid of refractive index 1.6, find the new apparent depth of the coin.
Hint
Apparent depth = 16 − 4 = 12 cm.
Answer
(a) $μ = \frac{16}{12} = \frac{4}{3}$ (b) Apparent depth $= \frac{16}{1.6} = 10$ cm (the coin now looks raised by 6 cm).
Q9 · 3 marks · Long answerHOTS / competency
A glass block 12 cm thick has a small air bubble inside. Viewed from the top face, the bubble appears 5 cm deep; viewed from the bottom face it appears 3 cm deep. Find the refractive index of the glass and the real distance of the bubble from the top face.
Hint
The two real depths add to 12 cm.
Answer
Sum of apparent depths $= 5 + 3 = 8$ cm $= \frac{12}{μ}$ , so $μ = 1.5$ . Real depth from the top $= 5 \times 1.5 = 7.5$ cm (and from the bottom $3 \times 1.5 = 4.5$ cm; $7.5 + 4.5 = 12$ ✓).
Q10 · 3 marks · Long answerHarder board level
State three factors on which the lateral displacement of a ray through a parallel-sided glass block depends, and how each affects it.
Hint
Thickness, angle, μ.
Answer
Thickness (thicker → more), angle of incidence (larger → more), refractive index / colour (higher μ, e.g. violet → more).
Q11 · 1 mark · MCQHOTS / competency
If red light is replaced by violet light at the same angle of incidence on a prism, the angle of deviation:
(a) increases
(b) decreases
(c) stays the same
(d) becomes zero
Hint
μ is larger for violet.
Answer
(a)
Q12 · 4 marks · Long answerHOTS / competency
A candle is placed in front of a thick glass mirror silvered at the back. (a) Why are several images seen? (b) Which image is the brightest, and why? (c) Would a thin mirror silvered at the front show this effect? Explain.
Hint
Partial reflection at the front glass surface.
Answer
(a) Light is partly reflected at the front glass surface and mostly at the silvered back; light reflected inside the glass is partly reflected again, giving several images. (b) The second image, formed by the silvered surface, which reflects most of the light. (c) No: with silvering at the front, light is reflected at one surface only, so only one image is formed.

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.State Snell's law.Show answer
sin i / sin r is constant for a given pair of media and colour of light.
2.Which quantity does not change on refraction?Show answer
Frequency.
3.Refractive index of glass and water?Show answer
About 1.5 and 1.33.
4.μ in terms of speeds?Show answer
μ = c/V
5.When does a ray pass undeviated?Show answer
When it falls normally on the surface.
6.Relation between real and apparent depth?Show answer
μ = real depth ÷ apparent depth.
7.At minimum deviation in a prism?Show answer
i = e and r₁ = r₂ = A/2.

Best order to study it

Refraction, cause, partial reflection.
Laws of refraction and refractive index.
Speed, wavelength, frequency; relative μ and reversibility.
Glass block and lateral displacement; thick mirror.
Real and apparent depth.
Prism: deviation and minimum deviation, ray diagrams.

Quick revision checklist

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

Can you state both laws of refraction?
Do you know what μ does and does not depend on?
Can you find speed and wavelength in a medium?
Can you use aμb × bμa = 1?
Can you draw a glass-block diagram with lateral displacement?
Can you draw a prism at minimum deviation with arrows?

Refraction of Light: frequently asked questions

Does the refractive index have a unit?

No. It is a ratio of two speeds (or of two sines).

Why does light bend at a boundary?

Because its speed changes; the part of the wavefront that enters first slows down (or speeds up) first, turning the ray.

Is the angle of the prism the same as the angle of deviation?

No. A is the angle between the two refracting faces; δ is the angle between the incident and emergent rays.

Why does a thick mirror form several images?

Light is partly reflected at the front glass surface and at the silvered back, and reflected again inside, giving several images; the second is brightest.

Are scale ray diagrams needed?

No. The syllabus asks for approximate ray diagrams with correct directions and angles marked.

Based on: CISCE ICSE 2027 Physics (Science Paper 1) syllabus; CISCE ICSE 2027 Physics (Science Paper 1) specimen question paper and marking scheme; CISCE Analysis of Pupil Performance, ICSE Physics (Science Paper 1), 2025; CISCE Analysis of Pupil Performance, ICSE Physics (Science Paper 1), 2024