What Is Total Internal Reflection
When light travels from an optically denser medium to a rarer medium, at the interface it is partly reflected back into the denser medium and partly refracted into the rarer one. The reflected part is called internal reflection. Because the second medium is rarer, the refracted ray bends away from the normal, so the angle of refraction is larger than the angle of incidence.
As the angle of incidence in the denser medium is increased, the angle of refraction increases faster. There comes an incidence angle at which the angle of refraction reaches its maximum possible value of $90^\circ$; the refracted ray then grazes along the interface. If the angle of incidence is increased even further, refraction is no longer possible and the entire incident ray is reflected back into the denser medium. This complete reflection is total internal reflection.
Total internal reflection is special because it loses no light. In ordinary reflection some fraction is always transmitted, so the reflected beam is weaker than the incident beam, however smooth the surface. In total internal reflection no transmission takes place at all and effectively one hundred per cent of the light is returned to the denser medium.
As incidence grows from the teal ray to the amber ray, refraction reaches $90^\circ$; the red ray beyond the critical angle is totally reflected back into the water.
Critical Angle and the Condition
The angle of incidence in the denser medium for which the angle of refraction in the rarer medium becomes $90^\circ$ is called the critical angle, written $i_c$. Applying Snell's law at the boundary between a denser medium of refractive index $n_1$ and a rarer medium of index $n_2$, with the angle of refraction set to $90^\circ$:
$$ n_1 \sin i_c = n_2 \sin 90^\circ \quad\Rightarrow\quad \sin i_c = \frac{n_2}{n_1} $$
When the rarer medium is air, $n_2 = 1$ and the denser medium has refractive index $n = n_1$, so this reduces to the form used in most NEET problems:
$$ \sin i_c = \frac{1}{n}, \qquad n = \frac{1}{\sin i_c} $$
For angles of incidence larger than $i_c$, the right-hand side of Snell's law would demand $\sin r > 1$, which is impossible. Snell's law cannot be satisfied, no refraction occurs, and the light is totally internally reflected. Two conditions must therefore hold together for total internal reflection, and both are mandatory.
| Condition | Requirement | If it fails |
|---|---|---|
| Direction of travel | Light must go from denser medium to rarer medium | Ray bends toward normal; TIR can never occur |
| Angle of incidence | $i > i_c$ in the denser medium | Ray refracts into the rarer medium (with partial reflection) |
TIR has a direction, and the boundary behaviour at $i = i_c$ is exact
Three errors recur in the exam. First, total internal reflection happens only when light goes denser → rarer; a ray entering a denser medium can never be totally internally reflected. Second, exactly at $i = i_c$ the refracted ray does not vanish, it grazes along the surface at $r = 90^\circ$ — total reflection begins only for $i > i_c$. Third, candidates invert the relation under pressure.
Remember $\sin i_c = 1/n$, so a higher refractive index gives a smaller critical angle.
Critical Angles of Common Media
The dependence of $i_c$ on refractive index is shown clearly by the standard NCERT table of critical angles measured with respect to air. Diamond, with the largest refractive index, has by far the smallest critical angle, which is the root of its optical behaviour.
| Substance | Refractive index $n$ | Critical angle $i_c$ (w.r.t. air) |
|---|---|---|
| Water | 1.33 | 48.75° |
| Crown glass | 1.52 | 41.14° |
| Dense flint glass | 1.62 | 37.31° |
| Diamond | 2.42 | 24.41° |
Reading down the table, as $n$ rises from $1.33$ to $2.42$ the critical angle falls from nearly $49^\circ$ to about $24^\circ$. A small critical angle means that even rays striking an internal face quite obliquely will exceed $i_c$ and reflect totally, which is exactly why dense materials trap and recirculate light so effectively.
TIR is a special case of Snell's law. If the bending of rays at a boundary still feels shaky, revise Refraction of Light before attempting critical-angle numericals.
Worked Numericals
Most TIR questions reduce to one of two moves: convert a refractive index into a critical angle, or convert speeds or times into a refractive index ratio and then into $\sin i_c$. The two worked examples below cover both routes.
The refractive index of glass is $1.52$. Find the critical angle for a glass–air interface. (NIOS Example 20.4)
Using $\sin i_c = 1/n = 1/1.52 \approx 0.658$, so $i_c \approx 42^\circ$. This is why a $45^\circ$ ray inside such glass — exceeding $42^\circ$ — undergoes total internal reflection in a totally reflecting prism.
Two media A and B are separated by a plane boundary. The speed of light is $1.5 \times 10^8\ \text{m/s}$ in A and $2.0 \times 10^8\ \text{m/s}$ in B. Find the critical angle for these two media.
The indices follow from $n = c/v$: $n_A = (3\times10^8)/(1.5\times10^8) = 2$ and $n_B = (3\times10^8)/(2\times10^8) = 1.5$. Light must go from the denser medium A to the rarer medium B, so $n_A \sin i_c = n_B \sin 90^\circ$, giving $\sin i_c = n_B/n_A = 1.5/2 = 0.75$, hence $i_c = \sin^{-1}(0.75)$.
Applications in Nature and Technology
NCERT §9.4.1 and NIOS §20.4.1 list four applications that recur in NEET as direct conceptual questions. Each one is simply the two conditions of total internal reflection met by a different geometry.
Mirage
On a hot day the air just above a road or desert sand is strongly heated, so the layers nearest the ground are less dense and have a lower refractive index than the cooler air above. A ray from a distant object such as a tree bends progressively as it passes downward through these layers, and when it meets a layer at an angle exceeding the critical angle for two consecutive layers, total internal reflection occurs. The eye traces the curved ray back to an inverted image below the object, producing the illusion of a pool of water.
Diamond sparkle
Diamond's refractive index of $2.42$ gives a critical angle of only $24.41^\circ$. Light entering a cut diamond strikes the internal faces at angles that almost always exceed this small value, so it undergoes repeated total internal reflection inside the crystal before finally emerging. This recirculation, combined with skilful cutting, is the physical origin of the brilliance of a diamond.
Optical fibres
An optical fibre is a hair-thin glass or quartz structure with a central core of higher refractive index surrounded by cladding of lower index. Light entering one end at a suitable angle strikes the core–cladding boundary at an angle larger than the critical angle, so it is totally internally reflected. It repeats this reflection thousands of times along the length and emerges at the far end with negligible loss of intensity — silica fibres can transmit more than 95% of the light over a length of one kilometre. Because each bounce is a total reflection, the signal survives even when the fibre is bent, which makes fibres ideal as light pipes for communications and for medical instruments that examine internal organs.
The ray strikes the core–cladding boundary above the critical angle at every point, so no light leaks into the cladding.
Totally reflecting prisms
A right-angled isosceles prism with angles $45^\circ$–$45^\circ$–$90^\circ$ is used to bend light by $90^\circ$ or by $180^\circ$, and to invert an image without changing its size. Light enters a face normally, strikes the hypotenuse face internally at $45^\circ$, and is totally reflected. For this to work the critical angle of the prism material must be less than $45^\circ$; the table shows that both crown glass ($41.14^\circ$) and dense flint glass ($37.31^\circ$) satisfy this, so a $45^\circ$ ray comfortably exceeds the critical angle.
The ray meets the hypotenuse at $45^\circ$, which exceeds the critical angle of glass, so it is totally reflected and leaves at right angles to its original path.
Total Internal Reflection in one screen
- TIR occurs only when light travels denser → rarer AND the angle of incidence exceeds the critical angle $i_c$.
- Critical angle: $\sin i_c = n_2/n_1$; for a medium of index $n$ in air, $\sin i_c = 1/n$ and $n = 1/\sin i_c$.
- At $i = i_c$ the refracted ray grazes the surface at $r = 90^\circ$; total reflection starts only for $i > i_c$.
- Higher refractive index ⇒ smaller critical angle (diamond $2.42 \Rightarrow 24.41^\circ$; water $1.33 \Rightarrow 48.75^\circ$).
- Applications: mirage, diamond sparkle, optical fibres (core $n$ > cladding $n$), and $45^\circ$ totally reflecting prisms (need $i_c < 45^\circ$).
- Unlike ordinary reflection, TIR returns essentially 100% of the light with no transmission.