PhysicsICSE Class 10 · Total Internal Reflection

Critical Angle: Definition, μ = 1/sin C and Solved Numericals (ICSE Class 10)

Q: What is the critical angle?

It is the angle of incidence, in the denser medium, at which light just grazes the boundary on leaving, so the angle of refraction in the rarer medium is 90°.

Q: How is the critical angle related to refractive index?

For a medium in air, sin C = 1/μ. A medium with a larger refractive index has a smaller critical angle.

Q: Why is the critical angle different for different colours?

The refractive index of a material is slightly different for each colour, greatest for violet and least for red. So violet has the smallest critical angle and red the largest.

Updated 15 October 2026 · Fact-checked

The critical angle C is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90°. For a medium in air, μ = 1/sin C. Glass (μ = 1.5) has C ≈ 42°, water C ≈ 48.6° and diamond C ≈ 24.4°; a larger μ gives a smaller C.

As i grows, the refracted ray bends toward the surface; at C it grazes (90°), beyond C all light reflects, and μ = 1/sin C.

Illustration drawn for QuestionBuddy.

Understand Critical Angle and μ = 1/sin C

From denser to rarer. Let a ray travel in glass towards a glass–air boundary, and let its angle of incidence $i$ grow steadily. Draw three sketches of the boundary with a dashed normal:
1. ** $i$ small: the ray refracts into air, bending away from the normal** ( $r > i$ ); a faint reflected ray also returns into the glass.
2. ** $i = C$ : the refracted ray just grazes the surface** ( $r = 90°$ ).
3. ** $i > C$ : there is no refracted ray at all; all the light is reflected back into the glass. This is total internal reflection** (TIR), with angle of reflection = angle of incidence.

Definition. The critical angle $C$ of a medium is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is $90°$ .

Relation with refractive index (derivation). Apply Snell's law for light going from the medium (μ) into air at the critical angle:

$\frac{s i n C}{s i n 90°} = \frac{1}{μ} .$

Since $sin 90° = 1$ :

$μ = \frac{1}{s i n C}, sin C = \frac{1}{μ} .$

Medium	μ	Critical angle
Water	1.33	about $48.6°$
Glass	1.5	about $42°$
Diamond	2.42	about $24.4°$

What the critical angle depends on.

The refractive index: a larger μ gives a smaller $C$ (diamond's tiny $C$ makes it sparkle).
The colour of light: μ is greatest for violet and least for red, so $C$ is smallest for violet and largest for red. A ray at a fixed angle may be totally reflected in violet light but escape in red light.
The second medium: for a boundary between a denser medium and a rarer one that is not air,

$sin C = \frac{μ _{rarer}}{μ _{denser}} .$

The closer the two refractive indices, the larger $C$ (glass–water has a larger critical angle than glass–air).

Temperature, slightly, through its effect on μ.

It does not depend on the angle of incidence; $C$ is a fixed property of the pair of media and the colour.

From speeds. Since $μ = \frac{c}{V}$ , $sin C = \frac{V _{denser}}{V _{rarer}}$ : the critical angle can be found from the two speeds of light.

Measuring C. Aim a narrow beam at the centre of the flat face of a semicircular glass block, entering through the curved face (along a radius, so it is not bent there). Increase the angle of incidence at the flat face until the refracted ray just grazes the surface; that angle is $C$ , and $μ = \frac{1}{s i n C}$ .

Critical Angle and μ = 1/sin C: formulas and key facts

Critical angle (medium in air): $μ = \frac{1}{s i n C}$; μ of the denser medium with respect to air.
Between two media: $sin C = \frac{μ _{rarer}}{μ _{denser}}$; Only from denser to rarer.
From speeds: $sin C = \frac{V _{denser}}{V _{rarer}}$; Because μ = c/V.

How to solve Critical Angle and μ = 1/sin C questions

Use sin C = 1/μ (or the two-media form), then compare C with the angle of incidence in the denser medium.

1Identify the denser medium (larger μ, or smaller speed of light).
2Find $sin C = \frac{1}{μ}$ , or $\frac{μ _{rarer}}{μ _{denser}}$ for two media, or $\frac{V _{denser}}{V _{rarer}}$ from speeds.
3Find $C$ from the sine (use the given table value or a calculator); if $C$ is given, $μ = \frac{1}{s i n C}$ .
4Measure the angle of incidence from the normal, not from the surface.
5Compare: $i > C$ gives TIR; $i < C$ gives refraction (with weak reflection); $i = C$ gives grazing emergence.

Critical Angle and μ = 1/sin C 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.

Why a diamond outshines glass

Suppose a ray inside a cut stone meets a face at an angle of incidence of $30°$ . Compare a real diamond (μ = 2.42) with a glass imitation (μ = 1.5).

In which stone is the ray totally reflected?

Try it first, then show the solutionSolution

Diamond: $sin C = \frac{1}{2.42}$ , $C \approx 24.4°$ ; $30° > 24.4°$ , so TIR.
Glass: $C \approx 41.8°$ ; $30° < 41.8°$ , so the ray escapes.

Answer: Only in the diamond; light bounces inside it many more times, so it sparkles more.

Memory hook: Bigger μ, smaller C, more sparkle.

Light from a pool lamp

Imagine a lamp on the floor of a swimming pool sends two rays up to the surface, meeting it at $40°$ and at $55°$ . (Critical angle of water $\approx 48.6°$ )

What happens to each ray?

Try it first, then show the solutionSolution

$40° < 48.6°$ : the ray refracts out into the air, bending away from the normal.
$55° > 48.6°$ : the ray is totally reflected back into the pool.

Answer: The $40°$ ray escapes; the $55°$ ray is totally reflected.

Memory hook: Steeper than C escapes; flatter than C stays in.

Tricks to remember Critical Angle and μ = 1/sin C

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

Sine C is one over μ
'C is the Sine of Shyness': the shy ray at C only grazes out, and sin C = 1/μ.
Use it when: Recalling the formula.
42, 48, 24
Glass 42°, water about 48.6°, diamond about 24°: 'forty-two glass, forty-eight wet, twenty-four sparkle'.
Use it when: Standard values.
Violet gives up first
Violet has the largest μ and the smallest C, so it is the first colour to be totally reflected as i increases.
Use it when: Colour-based TIR questions.
Three pictures
Small i: out and bent; i = C: along the surface; i > C: all back in.
Use it when: Explaining TIR with diagrams.

Critical Angle and μ = 1/sin C: solved examples

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

Example 1: Refraction at the critical angle · 1 mark · MCQBoard level

When light in a denser medium meets the boundary at the critical angle, the angle of refraction is:
(a) $0°$
(b) $45°$
(c) $90°$
(d) $180°$

Try it first, then show the solutionSolution

By definition, at $i = C$ the refracted ray grazes the surface, so $r = 90°$ .

Answer: (c) $90°$

Example 2: Critical angle of flint glass · 2 marks · Short answerBoard level

A dense flint glass has a refractive index of 1.6. Find its critical angle.

Try it first, then show the solutionSolution

$sin C = \frac{1}{μ} = \frac{1}{1.6} = 0.625$ . (1)
$C \approx 38.7°$ . (1)

Answer: About $38.7°$

Example 3: μ and speed from C · 2 marks · Short answerHarder board level

The critical angle of a transparent plastic is $37°$ . Taking $sin 37° = 0.60$ , find (a) its refractive index, (b) the speed of light in it. ( $c = 3 \times 1 0^{8}$ m/s)

Try it first, then show the solutionSolution

(a) $μ = \frac{1}{s i n C} = \frac{1}{0.60} \approx 1.67$ . (1)
(b) $V = \frac{c}{μ} = c sin C = 3 \times 1 0^{8} \times 0.60 = 1.8 \times 1 0^{8}$ m/s. (1)

Answer: (a) About 1.67 (b) $1.8 \times 1 0^{8}$ m/s

Example 4: Critical angle between two media · 3 marks · Short answerHarder board level

A glass of refractive index 1.6 is in contact with an oil of refractive index 1.44. (a) Calculate C at this glass–oil boundary. (b) What happens to rays in the glass meeting the boundary at $70°$ and at $60°$ ?

Try it first, then show the solutionSolution

(a) $sin C = \frac{1.44}{1.6} = 0.9$ , so $C \approx 64.2°$ . (1)
(b) $70° > 64.2°$ : total internal reflection. (1)
$60° < 64.2°$ : the ray refracts into the oil, bending away from the normal. (1)

Answer: (a) About $64.2°$ (b) $70°$ : TIR; $60°$ : refraction into the oil

Example 5: Critical angle from speeds · 3 marks · Short answerHOTS / competency

Light travels at $1.5 \times 1 0^{8}$ m/s in medium X and at $2.0 \times 1 0^{8}$ m/s in medium Y. (a) Which medium is optically denser? (b) Find the critical angle for the X–Y boundary. (c) In which direction of travel can TIR occur at this boundary?

Try it first, then show the solutionSolution

(a) X, because light is slower in it. (1)
(b) $sin C = \frac{V _{X}}{V _{Y}} = \frac{1.5}{2.0} = 0.75$ , so $C \approx 48.6°$ . (1)
(c) Only from X into Y (denser to rarer), at angles of incidence greater than $48.6°$ . (1)

Answer: (a) X (b) About $48.6°$ (c) From X to Y only

Example 6: Case study: measuring C with a semicircular block · 4 marks · Case-basedHOTS / competency

A student aims a narrow ray of red light through the curved face of a semicircular glass block towards the centre of its flat face. As she increases the angle of incidence at the flat face, the refracted ray just grazes the flat face when $i = 41.8°$ .
(a) Why is the ray not bent as it enters the curved face?
(b) Work out μ of this glass for red light.
(c) What happens when she increases $i$ to $45°$ ?
(d) With violet light, would the grazing angle be larger or smaller? Why?

Try it first, then show the solutionSolution

(a) Aimed at the centre, the ray travels along a radius, which is normal to the curved surface, so $i = 0°$ there. (1)
(b) $μ = \frac{1}{s i n 41.8°} \approx 1.5$ . (1)
(c) $45° > C$ : no refracted ray; the light is totally reflected back into the block. (1)
(d) Smaller: glass has a larger μ for violet, and $sin C = \frac{1}{μ}$ . (1)

Answer: (a) It meets the curved face normally (b) About 1.5 (c) TIR (d) Smaller

Where students lose marks in Critical Angle and μ = 1/sin C

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

Mistake: Reading the critical angle off the wrong angle in a diagram
The critical angle is measured from the normal, in the denser medium, for a refracted ray at 90°.
Wrong working
Taking the angle between the ray and the surface, or its supplement, as C.
Right working
Find the angle between the incident ray in the glass and the normal at the point where the ray grazes out.
Do this: Draw the normal first and mark angles from it.
Usually costs: Reported by CISCE examiners for 2024 Q4(iii), where candidates gave 47° or 133°.
Mistake: Choosing the wrong colour for TIR
A colour with a smaller critical angle is totally reflected more easily; that is the colour of larger μ (towards violet).
Wrong working
'Red light would undergo TIR instead of blue.'
Right working
'Violet (or indigo), whose critical angle is smaller than that of blue.'
Do this: Remember: violet has the largest μ and the smallest C.
Usually costs: Also reported for 2024 Q4(iii).
Mistake: Writing sin C = μ
μ > 1 but a sine can never exceed 1.
Wrong working
$sin C = 1.5$ .
Right working
$sin C = \frac{1}{1.5} \approx 0.667$ .
Do this: Check that your sine is less than 1.
Usually costs: The method mark.
Mistake: Measuring the angle of incidence from the surface
All angles in refraction are measured from the normal.
Wrong working
A ray making $30°$ with the surface is said to have $i = 30°$ .
Right working
$i = 90° - 30° = 60°$ .
Do this: Mark the normal before the angle.
Usually costs: Reported for 2025 Q4(i).

Exam tips

Learn the definition word for word: 'the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90°'.
Show the derivation $\frac{s i n C}{s i n 90°} = \frac{1}{μ}$ when asked for the relation; it earns the method mark.
In 2024 Q4(iii), the colour expected to undergo TIR in place of blue was one of shorter wavelength (towards violet), since its critical angle is smaller.
Give critical angles to one decimal place unless the question says otherwise.

Practice questions on Critical Angle and μ = 1/sin C

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 critical angle of a given glass depends on:
(a) the colour of the light
(b) the angle of incidence
(c) the thickness of the glass
(d) the brightness of the light
Hint
Through which quantity does μ change?
Answer
(a) the colour of the light
Q2 · 3 marks · Short answerHarder board level
With the help of three sketches, describe what happens to a ray going from glass into air as its angle of incidence increases from a small value to beyond the critical angle.
Hint
i < C, i = C, i > C.
Answer
(1) $i < C$ : the ray refracts into air, bending away from the normal, with a weak reflected ray. (2) $i = C$ : the refracted ray grazes the surface ( $r = 90°$ ). (3) $i > C$ : no refracted ray; the light is totally reflected back into the glass, with angle of reflection equal to $i$ .
Q3 · 1 mark · Very short answerBoard level
Find the critical angle of a medium of refractive index 1.8.
Hint
$sin C = \frac{1}{μ}$ .
Answer
$sin C = \frac{1}{1.8} \approx 0.556$ , so $C \approx 33.7°$ .
Q4 · 2 marks · Short answerHarder board level
For a certain liquid, $sin C = 0.64$ . Calculate μ for the liquid and how fast light travels through it. ( $c = 3 \times 1 0^{8}$ m/s)
Hint
$μ = \frac{1}{s i n C}$ , $V = c sin C$ .
Answer
$μ = \frac{1}{0.64} \approx 1.56$ ; $V = 3 \times 1 0^{8} \times 0.64 = 1.92 \times 1 0^{8}$ m/s.
Q5 · 2 marks · Short answerHarder board level
Medium A has μ = 1.65 and medium B has μ = 1.32. (a) Find the critical angle for the A–B boundary. (b) For which direction of travel is this critical angle meaningful?
Hint
Rarer over denser.
Answer
(a) $sin C = \frac{1.32}{1.65} = 0.8$ , so $C \approx 53.1°$ . (b) Only for light going from A (denser) into B (rarer).
Q6 · 2 marks · Short answerHarder board level
Arrange the critical angles of red, green and violet light in a glass prism in increasing order, and explain your order.
Hint
Relate μ to wavelength.
Answer
Violet < green < red. μ of glass is greatest for violet and least for red, and $sin C = \frac{1}{μ}$ , so the larger μ gives the smaller C.
Q7 · 4 marks · Long answerHOTS / competency
A ray in a glass block (μ = 1.5) meets the top surface at $45°$ . (a) Show that it is totally reflected if the top surface is in air. (b) A layer of oil (μ = 1.2) is now spread on the top surface. Show that the ray enters the oil, and find its angle of refraction in the oil. (c) The top of the oil layer is parallel to the glass and in contact with air. Does the ray escape into the air? Justify.
Hint
Find a critical angle for each boundary.
Answer
(a) Glass–air: $C \approx 41.8°$ , and $45° > 41.8°$ , so TIR. (b) Glass–oil: $sin C = \frac{1.2}{1.5} = 0.8$ , $C \approx 53.1°$ ; $45° < 53.1°$ , so it refracts: $sin r = \frac{1.5 s i n 45°}{1.2} \approx 0.884$ , $r \approx 62.1°$ . (c) No: it meets the oil–air surface at $62.1°$ , but the oil's critical angle is $sin^{- 1} \frac{1}{1.2} \approx 56.4°$ , so it is totally reflected there.

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.Relation between μ and C?Show answer
$μ = \frac{1}{s i n C}$ .
2.Critical angle of glass of μ 1.5?Show answer
About $42°$ .
3.Angle of refraction at the critical angle?Show answer
$90°$ .
4.Which colour has the smallest critical angle in glass?Show answer
Violet.
5.Does C depend on the angle of incidence?Show answer
No.

Critical Angle and μ = 1/sin C: frequently asked questions

What is the critical angle?

It is the angle of incidence, in the denser medium, at which light just grazes the boundary on leaving, so the angle of refraction in the rarer medium is 90°.

How is the critical angle related to refractive index?

For a medium in air, sin C = 1/μ. A medium with a larger refractive index has a smaller critical angle.

Why is the critical angle different for different colours?

The refractive index of a material is slightly different for each colour, greatest for violet and least for red. So violet has the smallest critical angle and red the largest.