Show that for a material with refractive index $\mu \geq \sqrt{2}$, light incident at any angle shall be guided along a length perpendicular to the incident face.

> [!TLDR] > For a material with refractive index $\mu \geq \sqrt{2}$, any ray entering one face will undergo total internal reflection at the adjacent perpendicular face, forcing it to travel along the length of the material — the critical angle condition becomes $\sin i_c \leq 1/\sqrt{2}$, which is always satisfied for any incident angle. The key insight here is **Total Internal Reflection (TIR)**. When light travels from a denser medium (the material, refractive index $\mu$) to a rarer medium (air, refractive index $1$), it bends away from the normal. If the angle of incidence inside the material exceeds a certain critical angle $i_c$, the ray cannot escape — it reflects back entirely, like a mirror. The problem asks: when does *any* ray entering one face get trapped and guided along a perpendicular direction? That means the ray must hit the adjacent face at an angle greater than $i_c$, no matter how it entered. Let’s set up the geometry. 1. **Define the path.** Imagine a rectangular slab. Light enters through the left face (the incident face). Inside the material, it bends toward the normal (since $\mu > 1$). It then travels to the top face (perpendicular to the incident face). For the ray to be guided along the length, it must reflect off this top face via TIR — so its angle of incidence at the top face must be $\geq i_c$. 2. **Relate the angles.** Let the angle of incidence at the left face be $i$ (measured from the normal to that face). Inside the material, the ray makes an angle $r$ with the normal, given by Snell’s law: $$1 \cdot \sin i = \mu \sin r \quad \Rightarrow \quad \sin r = \frac{\sin i}{\mu}.$$ Now, look at the top face. The normal to the top face is perpendicular to the normal of the left face. So the angle the ray makes with the top face’s normal is $90^\circ - r$. Call this angle $\theta$: $$\theta = 90^\circ - r.$$ For TIR at the top face, we need $\theta \geq i_c$, where $i_c$ is the critical angle for the material-air interfac

Show that for a material with refractive index μ ≥ √(2), light incident at any angle shall be guided along a l

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\mu \geq \sqrt{2}

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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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For a material with refractive index $\mu \geq \sqrt{2}$ , any ray entering one face will undergo total internal reflection at the adjacent perpendicular face, forcing it to travel along the length of the material — the critical angle condition becomes $\sin i_c \leq 1/\sqrt{2}$ , which is always satisfied for any incident angle.

The key insight here is Total Internal Reflection (TIR). When light travels from a denser medium (the material, refractive index $\mu$ ) to a rarer medium (air, refractive index $1$ ), it bends away from the normal. If the angle of incidence inside the material exceeds a certain critical angle $i_c$ , the ray cannot escape — it reflects back entirely, like a mirror.

The problem asks: when does any ray entering one face get trapped and guided along a perpendicular direction? That means the ray must hit the adjacent face at an angle greater than $i_c$ , no matter how it entered.

Let’s set up the geometry.

Define the path. Imagine a rectangular slab. Light enters through the left face (the incident face). Inside the material, it bends toward the normal (since $\mu > 1$ ). It then travels to the top face (perpendicular to the incident face). For the ray to be guided along the length, it must reflect off this top face via TIR — so its angle of incidence at the top face must be $\geq i_c$ .
Relate the angles. Let the angle of incidence at the left face be $i$ (measured from the normal to that face). Inside the material, the ray makes an angle $r$ with the normal, given by Snell’s law:

1 \cdot \sin i = \mu \sin r \quad \Rightarrow \quad \sin r = \frac{\sin i}{\mu}.

Now, look at the top face. The normal to the top face is perpendicular to the normal of the left face. So the angle the ray makes with the top face’s normal is $90^\circ - r$ . Call this angle $\theta$ :

\theta = 90^\circ - r.

For TIR at the top face, we need $\theta \geq i_c$ , where $i_c$ is the critical angle for the material-air interface:

\sin i_c = \frac{1}{\mu}.

Translate the condition. The condition $\theta \geq i_c$ becomes:

90^\circ - r \geq i_c \quad \Rightarrow \quad r \leq 90^\circ - i_c.

Taking sines (since all angles here are between $0^\circ$ and $90^\circ$ , sine is increasing):

\sin r \leq \sin(90^\circ - i_c) = \cos i_c.

But $\cos i_c = \sqrt{1 - \sin^2 i_c} = \sqrt{1 - \frac{1}{\mu^2}}$ .

Substitute for $\sin r$ . From Snell’s law, $\sin r = \frac{\sin i}{\mu}$ . So the TIR condition becomes:

\frac{\sin i}{\mu} \leq \sqrt{1 - \frac{1}{\mu^2}}.

Multiply through by $\mu$ :

$\sin i \leq \mu \sqrt{1 - \frac{1}{\mu^2}} = \sqrt{\mu^2 - 1}.$ …

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

COMEDK 2026Set 2026-A1 markMCQ
$42^{\circ}$
›Reveal solutionSolution
A ray entering normally into a prism of angle $45^\circ$ hits the second face at $45^\circ$ , which exceeds the critical angle of $42^\circ$ , so it undergoes total internal reflection — the correct option is (D).

The key here is to connect the prism angle, the path of the ray, and the critical angle condition. When a ray enters a prism normally (perpendicular to the surface), it does not bend at the first face. So the ray inside the prism travels straight toward the second face. The angle at which it meets that second face is simply the prism angle itself. If that angle is larger than the critical angle for the glass-air interface, the ray cannot escape — it reflects internally.

Why this approach works:

We don’t need Snell’s law at the first face because normal incidence means zero refraction. The only question is whether the ray can exit the second face. That depends on comparing the angle of incidence on the second face to the critical angle. The prism angle is given as $45^\circ$ , and the critical angle is $42^\circ$ . Since $45^\circ > 42^\circ$ , total internal reflection occurs.

Ray enters normally — The ray is perpendicular to the first face, so it passes straight into the glass without bending. Its direction inside the prism is exactly along the normal to the first face.

Geometry inside the prism — In a prism, the angle between the two faces is the prism angle $A = 45^\circ$ . The ray, traveling straight from the first face, will hit the second face at an angle equal to $A$ . This is because the ray’s path is perpendicular to the first face, and the second face is tilted by $A$ relative to the first. So the angle of incidence on the second face is $i = 45^\circ$ .

Compare with the critical angle — The critical angle for glass-air is given as $42^\circ$ . Total internal reflection occurs when the angle of incidence inside the denser medium exceeds the critical angle. Here $45^\circ > 42^\circ$ , so the ray cannot refract out — it must reflect internally. …
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COMEDK 2026Set 2026-M1 markMCQ
$15^{\circ}$
›Reveal solutionSolution

TIR needs light going from the denser medium (solid, $n=2.9$ ) to the rarer (liquid, $n=1.45$ ) at an angle $\ge$ the critical angle $30^{\circ}$ . The only choice in the correct direction and at that angle is "solid $\to$ liquid at $30^{\circ}$ ", listed as option (B).

Condition for total internal reflection. TIR can occur only when light travels from the optically denser medium into the rarer one, at an angle of incidence greater than or equal to the critical angle $\theta_c$ .

Here the solid ( $n_2=2.9$ ) is denser than the liquid ( $n_1=1.45$ ), so the ray must go solid $\to$ liquid. This immediately rules out any choice describing "liquid $\to$ solid" (rarer to denser can never give TIR).

Critical angle:

$\sin\theta_c=\frac{n_{\text{rarer}}}{n_{\text{denser}}}=\frac{1.45}{2.9}=0.5\ \Rightarrow\ \theta_c=30^{\circ}$ …
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KCET 2025Set D-41 markMCQ
$Q. The total energy carried by the light wave when it travels from a rarer to a non-reflecting and non-absorbing medium (A) remains same (B) increases (C) either increases or decreases depending upon angle of incidence (D) decreases$
›Reveal solutionSolution
Non-reflecting + non-absorbing means no energy is turned back and none is dissipated, so conservation of energy forces the transmitted energy to equal the incident energy.

Step 1 — Account for all the energy at an interface.

When light meets a boundary between two media, the incident energy in general splits three ways:
$E_{incident} = E_{reflected} + E_{absorbed} + E_{transmitted}$
Step 2 — Apply the stated conditions.

The question says the second medium is non-reflecting and non-absorbing. Therefore
$E_{reflected} = 0, \qquad E_{absorbed} = 0$
Substituting:
$E_{transmitted} = E_{incident}$
So the total energy carried by the wave is unchanged. There is simply no physical channel left through which it could grow or shrink — and energy cannot be created, so "increases" is ruled out on principle.

Step 3 — What does change on refraction, and what does not.

It is worth being precise, because this is where the distractors live:

Frequency $\nu$ — unchanged. The oscillating fields at the boundary drive the second medium at the same rate; the frequency is set by the source. This is why the photon energy $E = h\nu$ of each photon is also unchanged.

Speed $v$ — changes: $v = c/n$ .

Wavelength $\lambda$ — changes: $\lambda = v/\nu = \lambda_{vac}/n$ .

Intensity (energy per unit area per unit time) — can change, because the beam cross-section and speed change — but the total energy carried does not. …
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COMEDK 2025Set 2025-A1 markMCQ
$\frac{3}{2}$
›Reveal solutionSolution
The problem asks for the critical angle for total internal reflection when light goes from glass (n = 3/2) to water (n = 4/3). The minimum angle of incidence for no light entering water is the critical angle, given by $\sin i_c = n_{\text{water}} / n_{\text{glass}} = (4/3)/(3/2) = 8/9$ , so $i_c = \sin^{-1}(8/9)$ . The correct option is (B).

Concept and Intuition

Total internal reflection occurs when light travels from a denser medium (higher refractive index) to a rarer medium (lower refractive index) and the angle of incidence exceeds a certain threshold called the critical angle. At the critical angle, the refracted ray grazes the boundary (angle of refraction = 90°). For angles larger than this, no light enters the second medium — it is all reflected back. Here, glass (n = 1.5) is denser than water (n ≈ 1.333), so total internal reflection is possible. The minimum angle for which no light enters water is exactly this critical angle.

Step-by-step solution

Identify the refractive indices

Glass: $n_g = \frac{3}{2}$

Water: $n_w = \frac{4}{3}$

Since $n_g > n_w$ , light travels from denser to rarer medium.

Recall Snell’s law at the critical angle

Snell’s law: $n_g \sin i = n_w \sin r$

At the critical angle $i_c$ , the angle of refraction $r = 90^\circ$ , so $\sin r = 1$ .

Thus:

$n_g \sin i_c = n_w \cdot 1$

Solve for $\sin i_c$
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COMEDK 2025Set 2025-M1 markMCQ
$Q. Two light rays A and B travel from a medium into air at angles of incidence 15 degrees and 42 degrees respectively. In the medium, light travels 3 cm in 0.2 ns . Will there be total internal reflection? If so, which ray? (A) No (B) Yes, A (C) Yes, B (D) Yes, Both A and B$
›Reveal solutionSolution
The medium's speed gives $n=2$ , so the critical angle is $\theta_c=30^\circ$ . Ray A ( $15^\circ$ ) escapes, but ray B ( $42^\circ>30^\circ$ ) is totally internally reflected. The correct option is (C).

Total internal reflection occurs when light travels from a denser to a rarer medium at an angle of incidence exceeding the critical angle $\theta_c$ , where $\sin\theta_c=\dfrac{n_{\text{air}}}{n_{\text{medium}}}$ .

Speed of light in the medium. It covers $3\text{ cm}=0.03\text{ m}$ in $0.2\text{ ns}=0.2\times10^{-9}\text{ s}$ :

$v=\frac{0.03}{0.2\times10^{-9}}=1.5\times10^{8}\ \text{m/s}.$

Refractive index.

$n=\frac{c}{v}=\frac{3\times10^{8}}{1.5\times10^{8}}=2.0.$

Critical angle for the medium-to-air interface: …
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COMEDK 2024Set 2024-A1 markMCQ
$n$
›Reveal solutionSolution

The most demanding case is light entering at grazing incidence; requiring total internal reflection at the side wall then gives $n\ge\sqrt2$ , so $n_{min}=\sqrt2$ .

Light enters the flat end face at angle $i$ and refracts at $r$ , then strikes the lateral face at angle $(90^\circ-r)$ . For it never to escape the side, this must exceed the critical angle $\theta_c$ for all $i$ :

$90^\circ-r\ge\theta_c\ \Rightarrow\ \cos r\ge\sin\theta_c=\frac{1}{n}.$ …
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KCET 2023Set A-31 markMCQ
$\dfrac{I}{4}$
›Reveal solutionSolution

Unpolarised light loses half its intensity at the first polaroid; then apply Malus' law and solve $\tfrac12\cos^2\theta = \tfrac14$ .

1. First polaroid — the factor of one half.

Unpolarised light contains all vibration directions with equal probability. A polaroid transmits only the component along its pass axis, and averaging $\cos^2\theta$ over all angles gives $\langle\cos^2\theta\rangle = \tfrac12$ . Hence, for any orientation of the first polaroid:
$I_1 = \frac{I}{2}$
and this light is now plane-polarised along the first pass axis.

2. Second polaroid — Malus' law.

The light reaching the second polaroid is polarised, so Malus' law applies. Only the component of the electric field along the second pass axis gets through, $E_2 = E_1\cos\theta$ , and since intensity $\propto E^2$ :
$I_2 = I_1\cos^2\theta$
where $\theta$ is the angle between the pass axes.

3. Impose the given condition. …
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KCET 2021Set B-21 markMCQ
$\dfrac{3}{2}$
›Reveal solutionSolution
Apply the lens-maker's formula twice, using the refractive index of the glass relative to the surrounding medium each time; the ratio of the $(\mu_{\text{rel}}-1)$ terms gives $f_w : f_a = 4 : 1$ .

1. The governing formula
$\frac{1}{f} = \left(\mu_{\text{rel}} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
where $\mu_{\text{rel}} = \dfrac{\mu_{\text{lens}}}{\mu_{\text{medium}}}$ is the refractive index of the lens with respect to the medium it sits in. This is the crux of the problem: the radii are fixed by the glass, but the relative index changes when you move the lens from air to water.

Let the (constant, purely geometric) factor be
$K = \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$
2. Extract the absolute indices from the data

Air → glass: ${}_{a}\mu_{g} = \dfrac{3}{2} = 1.5 \quad\Rightarrow\quad \mu_{\text{glass}} = 1.5$ (taking $\mu_{\text{air}} = 1$ )

Air → water: ${}_{a}\mu_{w} = \dfrac{4}{3} \quad\Rightarrow\quad \mu_{\text{water}} = \dfrac{4}{3}$

3. Focal length in AIR
$\mu_{\text{rel}} = \frac{\mu_{\text{glass}}}{\mu_{\text{air}}} = \frac{3/2}{1} = \frac{3}{2}$ $\frac{1}{f_a} = \left(\frac{3}{2} - 1\right)K = \frac{1}{2}K \qquad(1)$
4. Focal length in WATER
$\mu_{\text{rel}} = {}_{w}\mu_{g} = \frac{\mu_{\text{glass}}}{\mu_{\text{water}}} = \frac{3/2}{4/3} = \frac{3}{2}\times\frac{3}{4} = \frac{9}{8} = 1.125$ $\frac{1}{f_w} = \left(\frac{9}{8} - 1\right)K = \frac{1}{8}K \qquad(2)$
5. Take the ratio

Divide (1) by (2):
$\frac{1/f_a}{1/f_w} = \frac{\tfrac{1}{2}K}{\tfrac{1}{8}K} = \frac{1}{2}\times\frac{8}{1} = 4$
$\Longrightarrow \quad \frac{f_w}{f_a} = 4 \quad\Longrightarrow\quad \boxed{f_w : f_a = 4 : 1}$ …
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COMEDK 2021Set 2021-B1 markMCQ
$Q.The brilliance of diamond is due to the fact that (A) Critical angle for diamond-air interface is high (B) Critical angle is high and refractive index is low (C) Critical angle is low and refractive angle is high (D) Its refractive index is low.$
›Reveal solutionSolution

Diamond's high refractive index gives a low critical angle ( $\sin\theta_c = 1/\mu$ ), so light is totally internally reflected many times before emerging — the source of its brilliance.

The critical angle is set by $\sin\theta_c = 1/\mu$ . Diamond's refractive index is very high ( $\mu \approx 2.42$ ), giving a low critical angle of about $24^\circ$ . Because so many internal rays exceed this angle, they undergo repeated total internal reflection and finally emerge concentrated, producing the characteristic sparkle. Thus brilliance requi …
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Q.Show that for a material with refractive index $\mu \geq \sqrt{2}$ , light incident at any angle shall be guided along a length perpendicular to the incident face.

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)

Method: Proving a Total-Internal-Reflection Condition Holds for "Any" Angle of Incidence

Steps

1. The governing formula

2. Extract the absolute indices from the data

3. Focal length in AIR

4. Focal length in WATER

5. Take the ratio

Q.Show that for a material with refractive index μ≥2\mu \geq \sqrt{2}μ≥2​, light incident at any angle shall be guided along a length perpendicular to the incident face.

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Q.Show that for a material with refractive index $\mu \geq \sqrt{2}$ , light incident at any angle shall be guided along a length perpendicular to the incident face.

Unlock everything free for 14 days