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Total internal reflection is one of the most testable ideas in Class 12 ray optics: a single formula, a clear condition, and applications from diamonds to the internet. It is also where students quietly lose marks, by using the wrong pair of refractive indices or measuring the angle from the surface instead of the normal. This post covers the condition, four worked problems, how optical fibres work, common mistakes and a practice routine.
When does total internal reflection happen?
The NCERT Class 12 Physics chapter on ray optics defines the critical angle as the angle of incidence for which the angle of refraction is 90°. From Snell's law, n₁ sin i = n₂ sin r, with r = 90°:
sin C = n₂/n₁
This only has a solution when n₂ < n₁, so two conditions must both hold:
Light goes from a denser medium (higher n) to a rarer one (lower n).
The angle of incidence, measured from the normal, is greater than C.
For a medium of index n in air, sin C = 1/n. NCERT's table of critical angles with respect to air gives 48.75° for water (n = 1.33), 41.14° for crown glass (1.52), 37.31° for dense flint glass (1.62) and 24.41° for diamond (2.42). The small critical angle of diamond is why light entering a cut diamond bounces around inside before it escapes.
The JEE (Main) 2026 syllabus lists "total internal reflection and its applications" under Unit 16, Optics. Check the current syllabus for your exam year.
Worked example 1: the outside medium matters
Find the critical angle for glass of n = 1.5 (a) in air and (b) under water, n = 1.33.
(a) sin C = 1/1.5 = 0.667, so C ≈ 41.8°.
(b) sin C = 1.33/1.5 = 0.887, so C ≈ 62.5°.
Putting glass in water raises the critical angle, so total internal reflection becomes harder. Many students write sin C = 1/1.5 in both cases. Always ask: what is on the other side of the boundary?
Worked example 2: the circle of light under water
A small lamp sits 2.0 m below a still water surface (n = 4/3). Light escapes only through a circle on the surface. Find its radius.
sin C = 3/4, so C ≈ 48.6°.
Rays hitting the surface at more than C are totally reflected, so the edge of the circle is where the ray meets the surface at exactly C.
From the geometry, r = h tan C. With sin C = 3/4, cos C = √7/4, so tan C = 3/√7 ≈ 1.134.
r = 2.0 × 1.134 ≈ 2.27 m.
A neater general form is r = h/√(n² − 1) = 2/√(7/9) = 6/√7 ≈ 2.27 m. A fish looking up sees the whole sky squeezed into this circle.
Worked example 3: why prisms make good mirrors
A right-angled isosceles prism turns a beam through 90° if the light inside meets the long face at 45° and is totally reflected. What refractive index is needed?
We need 45° > C, so sin C < sin 45° = 1/√2.
Since sin C = 1/n, this means n > √2 ≈ 1.414.
NCERT makes the same point: such prisms need a critical angle below 45°, which holds for both crown glass and dense flint glass. A water-filled prism (n = 1.33) would not work, because its critical angle is about 48.8°.
How do optical fibres trap light?
NCERT describes each fibre as a core surrounded by cladding, with the core's refractive index higher than the cladding's. Light sent in at a suitable angle undergoes repeated total internal reflections and comes out at the far end, even when the fibre bends. The chapter adds that in silica glass fibres more than 95% of the light can be transmitted over 1 km.
Worked example 4. A fibre has a core of n₁ = 1.50 and cladding of n₂ = 1.45. Find the critical angle at the core-cladding boundary, and the largest angle to the axis at which light can enter from air and still be trapped.
sin C = 1.45/1.50 = 0.967, so C ≈ 75.2°.
Inside the core, a ray at angle θᵣ to the axis meets the wall at 90° − θᵣ from the normal. Trapping needs 90° − θᵣ ≥ C, so θᵣ ≤ 14.8°.
At the entry face, Snell's law gives sin θ = n₁ sin θᵣ = 1.50 × sin 14.8° ≈ 0.384.
θ ≈ 22.6°. This equals √(n₁² − n₂²) = √(2.25 − 2.1025) ≈ 0.384, a quantity called the numerical aperture.
The acceptance-angle calculation goes beyond the NCERT text, but it uses only Snell's law twice and appears in JEE-style problems.
Who made fibres practical?
Glass was not always clear enough. The Nobel Prize press release for 2009 explains that in 1966 Charles Kao calculated how to transmit light over long distances through glass fibres: with the purest glass, signals could travel over 100 kilometres, compared with only 20 metres for the fibres of the 1960s. The first ultrapure fibre was made four years later, in 1970. Kao received half the 2009 prize "for groundbreaking achievements concerning the transmission of light in fibers for optical communication." His paper with George Hockham, Dielectric-fibre surface waveguides for optical frequencies (1966), analysed the losses that had to be overcome.
Common mistakes and how to avoid them
Wrong direction. Light going from air into glass can never be totally internally reflected. Check that n₁ > n₂.
Wrong pair of indices. Use n₂/n₁ for the two media at that boundary, not 1/n by habit.
Angle from the surface. Convert to the angle from the normal.
Forgetting partial reflection. Below C, some light is still reflected; total internal reflection means all of it is.
Mixing the axis angle and the wall angle in fibres. Draw the normal at the wall before writing any angle.
A practice routine for this week
Day 1: Six critical angle problems with different pairs of media, including two with liquids outside.
Day 2: Four geometry problems: the circle of light, a coin seen through a glass block, a prism turning light by 90° and by 180°.
Day 3: Three optical fibre problems, finding the critical angle and the acceptance angle.
Day 4: A timed mixed set from NCERT and past papers, combining total internal reflection with refraction at the entry face.
Seeing the reflected and refracted rays change as you vary the angle makes the critical angle much easier to remember. The Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read. For the rest of the chapter, see our guide to the lens maker's formula and lens combinations.
Try worked example 2 again with the lamp in a liquid of n = 1.6 and see how the circle shrinks.
