PhysicsCBSE Class 12 · Ray Optics and Optical Instruments

Total Internal Reflection Class 12: Critical Angle, Fibres and Prisms

Q: Why is it called 'total' internal reflection?

Because all the light is reflected back into the denser medium; none is refracted out. Ordinary reflection from glass or a mirror always lets some light through or absorbs some.

Q: Does the critical angle depend on the colour of light?

Yes. The refractive index is slightly larger for violet than for red, so the critical angle is slightly smaller for violet.

Q: Why are prisms used instead of mirrors in binoculars?

A right-angled prism totally reflects the light at its hypotenuse, giving almost 100% reflection, and has no silvered coating to tarnish or flake.

Q: Can total internal reflection happen at a glass-water boundary?

Yes, for light inside the glass, because glass has a higher refractive index than water. The critical angle is larger (about 62°) than for glass-air.

Q: Are mirage and diamond brilliance in the 2026-27 syllabus?

No. They are listed as not to be assessed. Total internal reflection and optical fibres are in the syllabus.

Updated 15 October 2026 · Fact-checked

Total internal reflection happens when light travelling from a denser to a rarer medium meets the boundary at an angle of incidence greater than the critical angle C, where sin C = 1/n (for a denser medium of refractive index n against air). All the light is then reflected back. It makes optical fibres and reflecting prisms work.

Understand Total Internal Reflection

Look up from under water in a swimming pool and the sky appears only through a bright circle above you; beyond it, the surface acts like a mirror. That mirror-like surface is total internal reflection.

From refraction to total reflection. When light goes from a denser medium (refractive index $n_{1}$ ) to a rarer one ( $n_{2} < n_{1}$ ), it bends away from the normal: Snell's law $n_{1} sin i = n_{2} sin r$ gives r > i. As i increases, r increases faster. At a particular angle of incidence, the critical angle C, the refracted ray grazes along the surface (r = 90°). Then $n_{1} sin C = n_{2}$ , so $sin C = \frac{n _{2}}{n _{1}}$ ; for a medium of index n against air, $sin C = \frac{1}{n}$ . For any angle of incidence larger than C, Snell's law would need sin r > 1, which is impossible: no light is refracted and all of it is reflected. This is total internal reflection (TIR).

The two conditions.
1. Light must travel from an optically denser to a rarer medium.
2. The angle of incidence must be greater than the critical angle for that pair of media.
Unlike ordinary reflection from a mirror (which always loses some light), TIR reflects essentially 100% of the light.

Critical angle values. Glass (n ≈ 1.5): C ≈ 42°. Water (n ≈ 1.33): C ≈ 49°. A larger refractive index means a smaller critical angle. Because n is slightly larger for violet than for red light, the critical angle is slightly smaller for violet.

Reflecting prisms. In a right-angled isosceles (45°-90°-45°) glass prism, light entering one short face normally meets the hypotenuse at 45°, which is more than the critical angle of glass (about 42°). It is totally reflected and leaves through the other short face, turned through 90°. Entering through the hypotenuse instead, it is totally reflected twice and turned through 180°. Such prisms replace mirrors in periscopes and binoculars because they reflect almost all the light and do not tarnish. The glass must have n > √2 ≈ 1.414 for this to work at 45°.

Optical fibres. A fibre has a thin glass or plastic core surrounded by a cladding of slightly lower refractive index. Light entering the end at a suitable angle strikes the core-cladding boundary at more than the critical angle and is totally reflected again and again, travelling along the fibre even when it bends, with very little loss. Uses: carrying telephone, internet and TV signals over long distances (light pipes), and in endoscopes that let doctors see inside the body. The angle of entry must be small enough for the ray to meet the boundary at more than C.

Escape cone. A point source at depth h under water lights up a circle on the surface of radius $r = h tan C$ ; rays outside this cone are totally reflected back into the water.

Total Internal Reflection: formulas and key facts

Critical angle (medium to air): $sin C = \frac{1}{n}$; C ≈ 42° for glass (n = 1.5), ≈ 49° for water.
Critical angle between two media: $sin C = \frac{n _{2}}{n _{1}} (n_{1} > n_{2})$; Light going from medium 1 to medium 2.
Snell's law: $n_{1} sin i = n_{2} sin r$; At i = C, r = 90°.
Escape circle above a submerged source: $r = h tan C = \frac{h}{n ^{2} - 1}$; h = depth of the source.
Reflecting prism condition: $n > \frac{1}{s i n 45°} = 2$; For TIR at 45° in a right-angled isosceles prism.
Largest entry angle for a fibre (from air): $sin i_{ma x} = n_{1}^{2} - n_{2}^{2}$; n₁ core, n₂ cladding.

How to solve Total Internal Reflection questions

Use this method for critical angle, prism and fibre questions.

1Check the direction: TIR is possible only going from the higher-n medium to the lower-n medium.
2Find C from $sin C = \frac{n _{2}}{n _{1}}$ (or $\frac{1}{n}$ against air).
3Find the actual angle of incidence at the boundary from the geometry (prism angles, normal incidence at entry, etc.).
4If the angle of incidence exceeds C, the light is totally reflected; otherwise part of it is refracted out.
5For a source under water, use $r = h tan C$ for the radius of the bright circle.
6For fibres, apply Snell's law at the entry face and the TIR condition at the core-cladding boundary.
7State the answer in degrees and say whether TIR occurs.

Total Internal Reflection 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.

Escape circle of a pool light

Suppose a small lamp is fixed at the bottom of a swimming pool, 1.5 m below the surface. Take the refractive index of water as 4/3.

(a) Find the critical angle for water. (b) Find the radius of the bright circle on the surface through which light leaves the water. (c) Find its area. What happens to rays striking the surface outside this circle?

Try it first, then show the solutionSolution

$sin C = \frac{3}{4}$ , so C ≈ 48.6°.
$r = h tan C = \frac{h}{n ^{2} - 1} = \frac{1.5}{7 /3} = \frac{4.5}{7} \approx 1.70$ m.
Area $= π r^{2} \approx 9.1$ m². Rays reaching the surface outside the circle meet it at more than C and are totally reflected back into the pool.

Answer: (a) about 48.6°; (b) about 1.70 m; (c) about 9.1 m²; rays outside are totally reflected.

Memory hook: Under water the whole sky squeezes into a circle; outside it, the surface is a mirror.

Prisms as perfect mirrors

Imagine building a periscope from two right-angled isosceles prisms instead of plane mirrors. You have glass of refractive index 1.5 and some plastic of refractive index 1.4.

(a) Show that the glass prisms reflect light totally at the hypotenuse. (b) Would plastic prisms work? (c) Why are prisms preferred to mirrors?

Try it first, then show the solutionSolution

Light entering a short face normally meets the hypotenuse at 45°. For glass, C = sin⁻¹(1/1.5) ≈ 41.8° < 45°, so TIR occurs.
For the plastic, C = sin⁻¹(1/1.4) ≈ 45.6° > 45°, so light at 45° is partly refracted out: the prism would not reflect totally.
TIR reflects essentially all the light, giving brighter images, and a prism has no reflecting coating to tarnish.

Answer: (a) 45° > 41.8°, so TIR; (b) no, 45° < 45.6°; (c) nearly 100% reflection and no tarnishing.

Memory hook: A 45° prism needs n above √2 to act as a perfect mirror.

Tricks to remember Total Internal Reflection

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

Dense to rare, beyond critical
Two conditions, in that order: Denser → Rarer, and i > C.
Use it when: Stating the conditions.
sin C = 1 over n
Bigger n, smaller C: diamond-like media trap light more easily than water.
Use it when: Comparing critical angles.
45 needs root 2
For TIR at 45°, n must exceed √2 ≈ 1.414.
Use it when: Prism questions.
Core high, cladding low
Light stays in the fibre because the core has the higher refractive index.
Use it when: Optical fibre questions.

Total Internal Reflection: solved examples

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

Example 1: Critical angle of glass · 2 marks · Very short answerBoard level

Find the critical angle for light going from glass of refractive index 1.5 into air. What happens to a ray inside the glass meeting the surface at 45°?

Try it first, then show the solutionSolution

$sin C = \frac{1}{1.5} = 0.667$ , C ≈ 41.8°. (1)
45° > 41.8°, so the ray is totally internally reflected. (1)

Answer: About 41.8°; the ray at 45° is totally reflected.

Example 2: Conditions and optical fibre · 3 marks · Short answerBoard level

State the two conditions for total internal reflection. Explain, with a diagram, how light travels along an optical fibre, and give two uses.

Try it first, then show the solutionSolution

Conditions: light must go from a denser to a rarer medium, and the angle of incidence must exceed the critical angle. (1)
A fibre has a core of higher refractive index and a cladding of lower index. Light entering the end strikes the core-cladding boundary at more than C and is totally reflected repeatedly, zigzagging along the fibre (diagram: core, cladding, zigzag ray). (1)
Uses: transmitting telephone and internet signals; endoscopes for viewing inside the body. (1)

Answer: Denser to rarer and i > C; light is guided by repeated TIR at the core-cladding boundary; used in communication and endoscopy.

Example 3: Right-angled prism · 3 marks · Short answerHarder board level

A ray enters normally through one face of a right-angled isosceles glass prism (n = 1.5). (a) Trace its path and find the total deviation. (b) Find the least refractive index for which this works. (c) What would happen with a prism of n = 1.3?

Try it first, then show the solutionSolution

(a) It enters undeviated and meets the hypotenuse at 45° > 41.8°, so it is totally reflected and leaves the other short face normally: deviation 90°. (1)
(b) Need 45° > C: $sin 45° > \frac{1}{n}$ , so $n > 2 \approx 1.414$ . (1)
(c) For n = 1.3, C ≈ 50.3° > 45°: the ray is mostly refracted out through the hypotenuse (with weak partial reflection). (1)

Answer: (a) deviated through 90° by TIR; (b) n > √2; (c) no TIR, light escapes through the hypotenuse.

Example 4: Source under water · 2 marks · Very short answerHarder board level

A point source lies 2 m below the surface of water (n = 4/3). Find the radius of the circle on the surface through which light emerges.

Try it first, then show the solutionSolution

$sin C = \frac{3}{4}$ ; $tan C = \frac{3}{7}$ . (1)
$r = h tan C = 2 \times \frac{3}{7} \approx 2.27$ m. (1)

Answer: About 2.27 m.

Example 5: Light in an optical fibre · 5 marks · Long answerHOTS / competency

An optical fibre has a core of refractive index 1.5 and a cladding of refractive index 1.4.
(a) Find the critical angle at the core-cladding boundary. (1)
(b) Light enters the flat end of the fibre from air at angle i to its axis. Show that it is guided by TIR only if $sin i \leq n_{1}^{2} - n_{2}^{2}$ , and find the largest i. (3)
(c) Why is the cladding needed at all? (1)

Try it first, then show the solutionSolution

(a) $sin C = \frac{1.4}{1.5} \approx 0.933$ , C ≈ 69.0°. (1)
(b) At entry: $sin i = n_{1} sin r$ . The ray then meets the side wall at angle $(90° - r)$ , which must be at least C: $cos r \geq sin C = \frac{n _{2}}{n _{1}}$ . (1)
So $sin r \leq 1 - \frac{n _{2}^{2}}{n _{1}^{2}}$ and $sin i = n_{1} sin r \leq n_{1}^{2} - n_{2}^{2}$ . (1)
$sin i_{ma x} = 2.25 - 1.96 \approx 0.539$ , so $i_{ma x} \approx 32.6°$ . (1)
(c) It protects the core surface and guarantees a lower-index boundary all along the fibre, so TIR is not spoilt by dirt, scratches or contact with other fibres. (1)

Answer: (a) about 69.0°; (b) i_max ≈ 32.6°; (c) to keep a clean lower-index boundary for TIR.

Example 6: Glass under water · 4 marks · Case-basedHarder board level

Read the passage and answer.

A glass block (n = 1.5) is placed at the bottom of a tank of water (n = 1.33). Light inside the glass strikes the glass-water boundary.
(a) Find the critical angle for the glass-water boundary. (1)
(b) Compare it with the glass-air critical angle. (1)
(c) A ray in the glass meets the boundary at 50°. Is it totally reflected (i) at a glass-air boundary, (ii) at the glass-water boundary? (1)
(d) Can light in the water be totally reflected at the water-glass boundary? (1)

Try it first, then show the solutionSolution

(a) $sin C = \frac{1.33}{1.5} \approx 0.887$ , C ≈ 62.5°. (1)
(b) Larger than 41.8° for glass-air, because the two media differ less. (1)
(c) (i) 50° > 41.8°: yes. (ii) 50° < 62.5°: no, it is partly refracted into the water. (1)
(d) No: water is the rarer medium relative to glass, so light going from water to glass bends towards the normal. (1)

Answer: (a) about 62.5°; (b) larger; (c) yes at glass-air, no at glass-water; (d) no.

Where students lose marks in Total Internal Reflection

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

Mistake: Stating only one condition for TIR.
Forgetting the direction (denser to rarer).
Do this: Always give both conditions.
Usually costs: ½ to 1 mark.
Mistake: Using sin C = n instead of 1/n.
Inverting the ratio.
Wrong working
sin C = 1.5
Right working
sin C = 1/1.5
Do this: sin C must be less than 1: rarer over denser.
Usually costs: The answer mark.
Mistake: Taking the angle of incidence on the hypotenuse as 90° or 0°.
Not using the prism geometry.
Do this: For normal entry into a 45°-90°-45° prism, the angle at the hypotenuse is 45°.
Usually costs: 1 mark.
Mistake: Using mirage or diamond brilliance as examples.
They appear in older books.
Do this: For 2026-27, use optical fibres and reflecting prisms; mirage and diamond brilliance are not in the syllabus.
Usually costs: Possibly the example mark.

Exam tips

Draw the three-ray diagram (i < C refracted, i = C grazing, i > C totally reflected) when explaining TIR.
Total internal reflection has carried 2- to 5-mark questions, including prism problems (for example 2023 55/1/1 Q35, 2024 55/5/1 Q30 and 2026 55/1/1 Q28).
In prism problems, mark every angle on the diagram before deciding whether TIR occurs.
When asked for uses, give optical fibres (with core and cladding) and reflecting prisms.
Remember C depends on colour: slightly smaller for violet than for red.

Practice questions on Total Internal Reflection

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 for a medium of refractive index $\frac{2}{3}$ is
(a) 30° (b) 45° (c) 60° (d) 90°
Hint
sin C = 1/n = √3/2.
Answer
(c) 60°.
Q2 · 1 mark · Very short answerBoard level
State two conditions necessary for total internal reflection.
Hint
Direction and angle.
Answer
Light must travel from a denser to a rarer medium, and the angle of incidence at the boundary must exceed the critical angle.
Q3 · 2 marks · Very short answerHarder board level
The critical angle for a transparent medium is 30°. Find its refractive index and the speed of light in it.
Hint
n = 1/sin C.
Answer
n = 1/sin 30° = 2; v = c/n = 1.5 × 10⁸ m/s.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Total internal reflection cannot take place when light goes from air into glass.
Reason (R): For total internal reflection light must travel from an optically denser to an optically rarer medium.
(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
Which way does the ray bend going into glass?
Answer
(a).
Q5 · 2 marks · Very short answerHOTS / competency
Light falls normally on one face of an equilateral glass prism. What is the least refractive index for which it is totally reflected at the second face?
Hint
The angle of incidence at the second face equals the prism angle.
Answer
The ray meets the second face at 60°. TIR needs 60° > C, i.e. n > 1/sin 60° = 2/√3 ≈ 1.155.
Q6 · 3 marks · Short answerHOTS / competency
A glass has refractive index 1.51 for red light and 1.53 for violet light. Find the critical angle for each. If white light inside the glass meets the surface at 41°, which colour emerges?
Hint
Compare 41° with each critical angle.
Answer
Red: C = sin⁻¹(1/1.51) ≈ 41.5°; violet: C = sin⁻¹(1/1.53) ≈ 40.8°. At 41°, violet exceeds its critical angle and is totally reflected, while red (41° < 41.5°) is partly refracted out: the emerging light is reddish.
Q7 · 2 marks · Very short answerHarder board level
Why are optical fibres able to carry light round bends with very little loss?
Hint
What fraction of light is reflected at each bounce?
Answer
Within the fibre the light meets the core-cladding boundary at angles greater than the critical angle (as long as the bend is gentle), so it is totally reflected at every bounce. TIR loses essentially no light, unlike a mirror, so the signal travels far with little loss.

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 critical angle and refractive index?Show answer
sin C = 1/n.
2.Critical angle for glass of n = 1.5?Show answer
About 41.8°.
3.Two conditions for TIR?Show answer
Denser to rarer, and angle of incidence greater than C.
4.Why is cladding of lower refractive index?Show answer
So light in the core is totally reflected at the core-cladding boundary.
5.Minimum n for TIR at 45°?Show answer
√2.
6.Deviation by a right-angled prism with normal entry on a short face?Show answer
90°.

Total Internal Reflection: frequently asked questions

Why is it called 'total' internal reflection?

Because all the light is reflected back into the denser medium; none is refracted out. Ordinary reflection from glass or a mirror always lets some light through or absorbs some.

Does the critical angle depend on the colour of light?

Yes. The refractive index is slightly larger for violet than for red, so the critical angle is slightly smaller for violet.

Why are prisms used instead of mirrors in binoculars?

A right-angled prism totally reflects the light at its hypotenuse, giving almost 100% reflection, and has no silvered coating to tarnish or flake.

Can total internal reflection happen at a glass-water boundary?

Yes, for light inside the glass, because glass has a higher refractive index than water. The critical angle is larger (about 62°) than for glass-air.

Are mirage and diamond brilliance in the 2026-27 syllabus?

No. They are listed as not to be assessed. Total internal reflection and optical fibres are in the syllabus.