(a) Figure 9.28 shows a cross-section of a 'light pipe' made of a glass fibre of refractive index $1.68$. The outer covering of the pipe is made of a material of refractive index $1.44$. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure. ![Cross-section of an optical fibre light pipe, showing the range of incidence angles for total internal reflection](/figures/ray-optics-and-optical-instruments/cross-section-optical-fibre-light-pipe-range-fig-9-28.svg) (b) What is the answer if there is no outer covering of the pipe?

(a) Figure 9.28 shows a cross-section of a 'light pipe' made of a glass fibre of refractive index 1.68. The ou

Exercises · 9.17

1.68

Cross-section of an optical fibre light pipe, showing the range of incidence angles for total internal reflection — Figure 9.28

(b) What is the answer if there is no outer covering of the pipe?

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Concept understanding — Total Internal Reflection

Total Internal Reflection: When Light Decides to Stay Home

Imagine you're running on a beach toward the water. On sand, you run fast. The moment you hit the water, your speed drops — the water "resists" more. If you run at a shallow angle toward the waterline, your legs will suddenly slow down, and your body will twist. That twist is refraction — light bending when it changes speed between two media.

Now imagine the reverse: you're swimming in the water, heading toward the shore. You're moving slower in water, and you want to get out onto the fast sand. If you approach the shore at a very shallow angle — almost parallel to the beach — you might never make it out. The sudden speed-up as you hit the sand could "reflect" you back into the water. That's the intuition for total internal reflection.

The Core Idea

Light normally passes from one transparent medium to another (say, from water to air) and bends away from the normal — because it speeds up. But if the angle of incidence in the slower medium is large enough, the light can't escape. It gets completely reflected back inside the first medium. No light transmits. That's total internal reflection.

Important

Total internal reflection (TIR) occurs only when light travels from a denser (slower) medium to a rarer (faster) medium, and the angle of incidence exceeds a critical value.

The Two Conditions (Memorise These)

For TIR to happen, both must be true:

Light must go from a denser medium to a rarer medium (e.g., glass → air, water → air, diamond → air).

Denser means higher refractive index ( $n$ ). Light slows down in a denser medium.
Angle of incidence ( $i$ ) must be greater than the critical angle ( $C$ ).

The critical angle is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is exactly $90^\circ$ .

The Critical Angle — The Tipping Point

Look at the diagram in your mind: a ray in water heading toward the surface. As you increase the angle of incidence, the refracted ray in air bends more and more away from the normal. At some specific angle $C$ , the refracted ray skims exactly along the surface — angle of refraction $= 90^\circ$ .

\sin C = \frac{n_{\text{rarer}}}{n_{\text{denser}}}

For water ( $n = 1.33$ ) to air ( $n = 1.00$ ):

\sin C = \frac{1.00}{1.33} \approx 0.75 \quad\Rightarrow\quad C \approx 48.6^\circ

So if you shine a light from water into air at an angle greater than about $49^\circ$ from the normal, the light will not leave the water at all. It reflects back down — perfectly.

What Actually Happens at the Boundary?

$i < C$ : Most light refracts out; a little reflects (normal partial reflection).
$i = C$ : Refracted ray grazes the surface; transmitted intensity is nearly zero.
$i > C$ : No transmitted ray. All the light energy reflects back into the denser medium. The reflection is 100% — no absorption, no transmission.

Watch out

TIR is not the same as ordinary reflection from a mirror. In TIR, there is no silvering or coating. The reflection happens because the wave cannot exist in the rarer medium — it's forced back. This gives perfect reflection with zero energy loss, unlike a metal mirror which absorbs some light.

--- …

Why this formula?

Total Internal Reflection: Why the Key Formulas Hold

Total Internal Reflection (TIR) is a fascinating optical phenomenon where light, instead of escaping from a denser medium into a rarer one, gets completely reflected back into the denser medium. Let's build the understanding from first principles.

1. The Foundation: Snell's Law

The entire story begins with Snell's Law:

n_1 \sin \theta_1 = n_2 \sin \theta_2

Where:

$n_1$ = refractive index of the denser medium (e.g., glass, water)
$n_2$ = refractive index of the rarer medium (e.g., air)
$\theta_1$ = angle of incidence (in denser medium)
$\theta_2$ = angle of refraction (in rarer medium)

Key fact: $n_1 > n_2$ (light travels from denser to rarer).

2. The Critical Angle: Where Refraction "Bends" to 90°

As $\theta_1$ increases, $\theta_2$ increases faster (because $n_1 > n_2$ ). At some special angle, $\theta_2$ becomes exactly $90^\circ$ — the refracted ray grazes the surface.

Set $\theta_2 = 90^\circ$ in Snell's Law:

n_1 \sin \theta_c = n_2 \sin 90^\circ

Since $\sin 90^\circ = 1$ :

\boxed{\sin \theta_c = \frac{n_2}{n_1}}

Why this formula?

It's not arbitrary — it's the limit of Snell's Law. The critical angle $\theta_c$ is the largest incidence angle for which refraction is still possible. Beyond this, Snell's Law would demand $\sin \theta_2 > 1$ , which is impossible — no real angle satisfies it.

3. Beyond the Critical Angle: Why TIR Occurs

When $\theta_1 > \theta_c$ :

Snell's Law gives $\sin \theta_2 = \frac{n_1}{n_2} \sin \theta_1 > 1$
No real $\theta_2$ exists
Physics says: the wave cannot "fit" into the rarer medium
Result: All energy is reflected back into the denser medium

This isn't a failure of Snell's Law — it's a physical boundary where the wave's behaviour changes from propagating to evanescent (decaying).

4. The Condition for TIR (Exam-Ready Summary)

For Total Internal Reflection to occur, both conditions must hold:

Light travels from denser to rarer medium ( $n_1 > n_2$ ) …

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Total internal reflection needs the ray to hit the fibre wall at $\ge$ the critical angle. This sets a maximum entry angle with the axis: about $60^\circ$ with cladding, and $90^\circ$ (i.e. any ray) with no cladding.

Concept understanding

At the core-wall, TIR occurs when the angle of incidence (from the normal to the wall) is at least the critical angle $\theta_c$ , where $\sin\theta_c=n_2/n_1$ , with $n_1=1.68$ the core index and $n_2$ the outer medium. If the refracted ray makes angle $r$ with the axis, it meets the wall at $(90^\circ-r)$ from the wall's normal, so TIR requires $90^\circ-r\ge\theta_c$ , i.e. $r\le 90^\circ-\theta_c$ . Snell's law at the flat entrance face ( $n=1$ to $n_1$ ) then relates $r$ to the entry angle $i$ : $\sin i=1.68\sin r$ .

Part (a): with cladding ( $n_2=1.44$ )

\sin\theta_c=\frac{1.44}{1.68}=0.857\ \Rightarrow\ \theta_c\approx 59^\circ.

So $r\le 90^\circ-59^\circ=31^\circ$ . Maximum entry angle:

\sin i_{max}=1.68\sin 31^\circ\approx1.68\times0.515\approx0.865\ \Rightarrow\ i_{max}\approx 60^\circ.

Every ray entering within $0^\circ$ to $60^\circ$ of the axis is trapped.

Part (b): no cladding (air, $n_2=1.00$ )

$\sin\theta_c=\frac{1.00}{1.68}=0.595\ \Rightarrow\ \theta_c\approx 36.5^\circ,$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AHSEC Higher Secondary (HS) Final Examination 2025Set ANNUAL1 mark
$Q.Mention the condition where Snell's law of refraction cannot be satisfied.$
›Reveal solutionSolution

Snell's law cannot be satisfied beyond the critical angle, when going from denser to rarer medium — total internal reflection takes over.

Snell's law: n1 sinθ1 = n2 sinθ2, so sinθ2 = (n1/n2) sinθ1.

When light travels from a denser medium (n1) to a rarer medium (n2 < n1), (n1/n2) > 1. As the angle of incidence θ1 increases, at some angle called the critical angle θc, sinθ2 becomes exactly 1 (θ2 = 90°, the refracted ray grazes the surface).

…

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AHSEC Higher Secondary (HS) 1st Year Examination 2025Set ANNUAL1 markMCQ
$Q. Optical fibre is based on which of the following?$
(a) Total internal reflection
(b) Refraction
(c) Diffraction
(d) Polarization
›Reveal solutionSolution

Optical fibres work on the principle of total internal reflection (TIR) — a Class 12 Ray Optics topic, out of scope for the Class 11 chapter list used here.

An optical fibre consists of a thin core of high refractive index glass surrounded by a cladding of lower refractive index. Light entering the fibre strikes the core-cladding interface at an angle greater than the critical angle for that pair of media, so instead of refracting out, it undergoes total internal reflection and travels down the fibre by repeated TIR at the walls, with almost no loss of intensity, even around bends.

…

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AHSEC Higher Secondary (HS) Final Examination 2024Set ANNUAL1 mark
$Q.If the critical angle of water with respect to air is 48.75 and sin 48.75=0.75, cos 48.75=0.65 and tan 48.75=1.14 approximately, what will be the refractive index of water?$
›Reveal solutionSolution

Refractive index of water $n = 1/\sin C \approx 1.33$ .

At the critical angle $C$ , light travelling from the denser medium (water) to the rarer medium (air) refracts at $90^\circ$ . By Snell's law applied at the water-air interface:

$n \sin C = 1 \times \sin 90^\circ = 1$

$n = \frac{1}{\sin C}$

…

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AHSEC Higher Secondary (HS) Final Examination 2023Set ANNUAL1 markMCQ
$Q. Mirage is an optical phenomenon related to (Choose the correct option)$
(i) scattering
(ii) total internal reflection
(iii) total internal refraction
›Reveal solutionSolution

A mirage is caused by total internal reflection of light within air layers whose refractive index varies continuously with temperature.

On a hot day, the air just above a road or a desert surface is much hotter — and hence less dense, with a lower refractive index — than the air higher up. Light coming from the sky travels from a denser (cooler) layer towards progressively rarer (hotter) layers as it approaches the ground. At some layer, the angle of incidence exceeds the critical angle for that pair of layers, and the ray undergoes total internal reflection, bending back upward before it reaches the ground.

…

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AHSEC Higher Secondary (HS) Final Examination 2020Set ANNUAL1 mark
$Q. Name the equipment which can transmit optical signal through it and are used as 'light pipe'.$
›Reveal solutionSolution

Optical fibres act as 'light pipes' by trapping light inside a thin glass core through repeated total internal reflection.

An optical fibre consists of a thin cylindrical core of glass (or quartz) of high refractive index, surrounded by a cladding of lower refractive index. When light enters one end of the fibre at an angle greater than the critical angle for the core–cladding interface, it undergoes total internal reflection repeatedly at the core–cladding boundary as it travels down the fibre, so the light is guided along the length of the fibre with very little loss — effectively 'piping' the light from one end to the other, even around gentle bends. This makes optical fibres ex …

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AHSEC Higher Secondary (HS) Final Examination 2019Set ANNUAL1 mark
$Q. The sparkle of a diamond can be explained by which phenomenon of light?$
›Reveal solutionSolution

A diamond sparkles because its high refractive index gives it a very small critical angle, so light entering it undergoes multiple total internal reflections before exiting.

Diamond has a very high refractive index (about 2.42), so its critical angle $\theta_c = \sin^{-1}(1/n) \approx 24.4°$ is very small. Diamonds are faceted by jewellers so that light entering the top face strikes the internal sloped facets at angles greater than this small critical angle. At each such facet the light undergoes total internal reflection instead of escaping, bouncing repeatedly inside the stone until it eventually emerges from the top, concentrated and dispersed into its spectral colours. This trapping-and-bouncing of light, possible onl …

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Concept understanding — Total Internal Reflection

Total Internal Reflection: When Light Decides to Stay Home

The Core Idea

The Two Conditions (Memorise These)

The Critical Angle — The Tipping Point

What Actually Happens at the Boundary?

Why this formula?

Total Internal Reflection: Why the Key Formulas Hold

1. The Foundation: Snell's Law

2. The Critical Angle: Where Refraction "Bends" to 90°

3. Beyond the Critical Angle: Why TIR Occurs

4. The Condition for TIR (Exam-Ready Summary)

Concept understanding

Part (a): with cladding ( $n_2=1.44$ )

Part (b): no cladding (air, $n_2=1.00$ )

Method: Total Internal Reflection in an Optical Fibre (Angle-with-Axis Problems)

Steps

Concept understanding

Part (a): with cladding (n2=1.44n_2=1.44n2​=1.44)

Part (b): no cladding (air, n2=1.00n_2=1.00n2​=1.00)

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Part (a): with cladding ( $n_2=1.44$ )

Part (b): no cladding (air, $n_2=1.00$ )

Unlock everything free for 14 days