At what angle should a ray of light be incident on the face of a prism of refracting angle $60^\circ$ so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is $1.524$.

At what angle should a ray of light be incident on the face of a prism of refracting angle 60^° so that it jus

Exercises · 9.21

60^\circ

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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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The ray must enter the prism at an angle of incidence such that after refraction, it strikes the second face at exactly the critical angle. For a $60^\circ$ prism with $n = 1.524$ , the required angle of incidence is approximately $29.75^\circ$ .

The key idea here is that "just suffers total internal reflection" means the ray hits the second face at exactly the critical angle — any less and it would refract out, any more and it would still reflect but we want the threshold. This is a classic prism problem that ties together Snell's law at the first face and the critical angle condition at the second face, linked by the prism's geometry.

Let’s unpack why this works. Total internal reflection (TIR) happens when light tries to go from a denser medium (prism, $n = 1.524$ ) to a rarer medium (air, $n = 1$ ) at an angle greater than the critical angle. The critical angle $C$ is given by $\sin C = \frac{1}{n}$ . For the ray to "just" suffer TIR, the angle of incidence on the second face must equal $C$ . The prism angle $A = 60^\circ$ then relates this internal angle to the angle of refraction at the first face through simple geometry.

Watch out

A common mistake is to confuse the angle of incidence on the first face with the angle inside the prism. Remember: the ray bends at entry, so the internal angle is not the same as the external incidence angle.

Now let’s work through it step by step.

Find the critical angle for the prism material. The refractive index of the prism relative to air is $n = 1.524$ . For TIR at the second face, the angle of incidence inside the prism on that face must be at least the critical angle $C$ , where

\sin C = \frac{1}{n} = \frac{1}{1.524}

C = \sin^{-1}\left(\frac{1}{1.524}\right)

Calculating: $\frac{1}{1.524} \approx 0.6562$ , so $C \approx 41.0^\circ$ (more precisely, $41.0^\circ$ is a good approximation; let's keep it exact for now).

Relate the prism geometry to the internal angles. In a prism, the sum of the angles inside the triangle formed by the ray and the two faces is $180^\circ$ . Specifically, if $r_1$ is the angle of refraction at the first face (inside the prism) and $r_2$ is the angle of incidence on the second face (also inside the prism), then for a prism of angle $A$ :

r_1 + r_2 = A

This is because the normals at the two faces are inclined at angle $A$ , and the ray's path inside forms a triangle with them.

Here $A = 60^\circ$ , and for "just" TIR, $r_2 = C \approx 41.0^\circ$ . Therefore:

r_1 = A - r_2 = 60^\circ - C

So $r_1 \approx 60^\circ - 41.0^\circ = 19.0^\circ$ .

Apply Snell's law at the first face to find the external angle of incidence $i$ . At the first face, light goes from air ( $n = 1$ ) into the prism ( $n = 1.524$ ). Snell's law:

\sin i = n \sin r_1

Substituting $r_1 = 60^\circ - C$ :

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Here are the most common mistakes students make when solving this Total Internal Reflection (TIR) problem, along with how to avoid each.

Mistake 1: Confusing the angle of incidence at the first face with the angle at the second face

The error:

Students often think the given condition (“just suffers TIR at the other face”) refers to the angle of incidence at the first face. They then try to directly apply Snell’s law at the first surface using $1.524$ and $60^\circ$ , leading to a wrong answer.

How to avoid:

Read carefully: The TIR happens at the second face (the other face).
The angle of incidence at the second face ( $r_2$ ) must equal the critical angle for the prism material.
First find the critical angle $C$ using:

\sin C = \frac{1}{\mu} = \frac{1}{1.524}

C \approx 41^\circ

This $C$ is the value of $r_2$ at the second face.

Mistake 2: Forgetting the prism geometry relation between $r_1$ and $r_2$

The error:

After finding $r_2 = C$ , students sometimes forget that for a prism, the sum of the two refraction angles equals the prism angle $A$ :

r_1 + r_2 = A

They might incorrectly set $r_1 = A$ or $r_2 = A$ .

How to avoid:

Always write the prism relation first:

r_1 + r_2 = A

Here $A = 60^\circ$ and $r_2 = C \approx 41^\circ$ , so:

r_1 = 60^\circ - 41^\circ = 19^\circ

Mistake 3: Using the wrong Snell’s law sign or direction at the first face

The error:

When applying Snell’s law at the first face, students sometimes write:

\mu = \frac{\sin i}{\sin r_1}

but forget that the ray is going from air ( $\mu = 1$ ) into prism ( $\mu = 1.524$ ). They may invert the fraction.

How to avoid:

Always write Snell’s law in the correct order:

\mu_{\text{air}} \sin i = \mu_{\text{prism}} \sin r_1

Since $\mu_{\text{air}} = 1$ , this becomes:

\sin i = 1.524 \times \sin 19^\circ

Calculate $\sin 19^\circ \approx 0.3256$ , then:

\sin i \approx 1.524 \times 0.3256 \approx 0.496

i \approx \sin^{-1}(0.496) \approx 29.75^\circ

Mistake 4: Not checking if the calculated $i$ is physically possible

The error:

Students sometimes get $\sin i > 1$ due to a calculation slip (e.g., using $r_1 = 41^\circ$ instead of $19^\circ$ ). They then report an impossible angle.

How to avoid:

Always check: $\sin i$ must be $\leq 1$ . …

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

Showing the 12 most recent of 37 on this concept.

CBSE 2026Set 55/3/11 markMCQ
$Q. In which of the following does total internal reflection NOT occur ? (A) Twinkling of stars (B) Brilliance of diamonds (C) Optical fibre (D) Reflecting prism$
›Reveal solutionSolution
Total internal reflection requires light to travel from a denser to a rarer medium at an angle greater than the critical angle. Twinkling of stars involves refraction through atmospheric layers of varying density, not TIR. The answer is (A).

Total internal reflection is a phenomenon that occurs when light traveling in a denser medium strikes the boundary with a rarer medium at an angle greater than the critical angle. At this point, instead of refracting into the second medium, all the light reflects back into the denser medium. The key conditions are: light must go from higher to lower refractive index, and the angle of incidence must exceed the critical angle $\theta_c = \sin^{-1}(n_2/n_1)$ where $n_1 > n_2$ .

Let's examine each option to see where this principle applies:

Twinkling of stars: Starlight enters Earth's atmosphere and passes through layers of air with slightly different temperatures and densities. As light moves through these layers, it undergoes refraction — bending at each boundary because the refractive index changes gradually. Sometimes light bends toward regions of higher density, sometimes toward lower density, but it continues to propagate through the atmosphere. The random fluctuations in atmospheric density cause the apparent position and brightness of stars to change rapidly, creating the twinkling effect. This is purely refraction, not total internal reflection.

Brilliance of diamonds: A diamond has a very high refractive index ( $n \approx 2.42$ ). When light enters a diamond and strikes the internal faces, it often does so at angles greater than the critical angle (which is quite small, around $24.4°$ for diamond-air interface). The light then undergoes total internal reflection multiple times inside the diamond before emerging, creating the characteristic sparkle and brilliance. …
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CBSE 2026Set V11 markMCQ
$Q. For total internal reflection of light :$
(a) light should be travelling from rarer medium to denser medium
(b) light should be travelling from denser medium to rarer medium
(c) light should be incident along the normal
(d) angle of incidence should be equal to $90^\circ$
›Reveal solutionSolution

(b) light should be travelling from denser medium to rarer medium …

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CBSE 2026Set DS1 markMCQ
$\alpha$
i) $\beta < \alpha$
ii) $\beta > \alpha$
iii) $\beta = \alpha$
iv) $\beta \le \alpha$
›Reveal solutionSolution

TIR occurs only for angles of incidence larger than the critical angle, so $\beta > \alpha$ .

Concept. When light travels from a denser to a rarer medium, at the critical angle $\alpha$ the refracted ray just grazes the surface (angle of refraction $=90^\circ$ ). If the angle of incidence is increased beyond this critical angle, refraction is no longer possible and the light is completely reflected back into the denser medium — this is total internal reflection.

…

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CBSE 2026Set ANNUAL1 mark
$Q.When light passes from a denser medium to a rarer medium, and the angle of incidence equals the critical angle, what is the angle of refraction?$
›Reveal solutionSolution

The critical angle is DEFINED as the angle of incidence at which the refracted ray just grazes the interface, making the angle of refraction exactly 90 degrees.

When light travels from a denser medium towards a rarer medium, it bends away from the normal. As the angle of incidence is increased, the angle of refraction increases faster and eventually reaches 90 degrees (the refracted ray travels right along the boundary between the two media) - the angle of incidence at which this happens is called the critical angle (theta_c). By …

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CBSE 2026Set ANNUAL1 mark
$Q. Answer in one word/sentence: What is the phenomena that causes a bubble in water to shine brightly?$
›Reveal solutionSolution

Air bubbles in water appear to shine brightly (like silver) due to total internal reflection of light at the water-air interface of the bubble.

Light travelling in the denser medium (water) strikes the surface of the air bubble (a rarer medium) at angles greater than the critical angle for the water-air interface. Since it cannot refract out, it is totally internal …

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CBSE 2025Set JS1 mark
$Q. What do you mean by total internal reflection? Show it by drawing ray diagram.$
›Reveal solutionSolution
When light travelling in a denser medium hits a rarer medium at an angle greater than the critical angle, it is fully reflected back — this is total internal reflection.

Definition. When light passes from an optically denser medium to an optically rarer medium, it bends away from the normal. As the angle of incidence increases, the refracted ray bends more, until at the critical angle $\theta_c$ the refracted ray grazes the surface ( $r=90^\circ$ ). For any angle of incidence greater than $\theta_c$ , no refraction occurs and the light is completely reflected back into the denser medium. This is called total internal reflection (TIR).

Conditions.

Light must travel from a denser to a rarer medium.

The angle of incidence must exceed the critical angle: $i>\theta_c$ , where $\sin\theta_c = \dfrac{1}{n}$ ( $n$ = refractive index of the denser medium relative to the rarer).

Ray diagram (description). For a ray striking the denser–rarer boundary at three increasing angles:

$i<\theta_c$ : refracts into the rarer medium (bends away from normal).

$i=\theta_c$ : refracted ray travels along the surface ( $90^\circ$ ). …
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CBSE 2025Set D1 markMCQ
$Q. If the critical angle for total internal reflection from any medium to vacuum is 30°, then the velocity of light in the medium is (A) 3 \times 10^8 m/sec (B) 1.5 \times 10^8 m/sec (C) 6 \times 10^8 m/sec (D) 4.5 \times 10^8 m/sec$
›Reveal solutionSolution

From the critical angle, refractive index n = 1/sin C = 2; the speed of light in the medium is c/n = 1.5×10⁸ m/s.

The critical angle C for total internal reflection from a medium into vacuum is related to the refractive index n by

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

With C = 30°:

$n = \frac{1}{\sin 30^\circ} = \frac{1}{0.5} = 2$

…

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CBSE 2025Set A1 mark
$Q.Fill in the blank with appropriate word: In total internal reflection, refractive angle is ______, when incidence angle is equal to the critical angle.$
›Reveal solutionSolution

At the critical angle, the refracted ray just grazes the boundary, so the angle of refraction is 90°.

The critical angle C is defined, for light travelling from a denser to a rarer medium, as that particular angle of incidence for which the angle of refraction becomes exactly 90°, i.e. the refracted ray travels along the boundary/interface separating the two media. By Snell's law, n1 sin C = n2 sin 90° = n2, which gives sin C = n2/n1. For any …

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CBSE 2025Set ANNUAL1 mark
$Q.How is the critical angle related to the refractive index?$
›Reveal solutionSolution

The critical angle is the angle of incidence (in the denser medium) at which the refracted ray just grazes the interface at 90 degrees, which by Snell's law gives sin(C) = 1/n.

Consider light travelling from a denser medium (refractive index n, relative to a rarer medium like air) toward the interface. By Snell's law, for the boundary case where the angle of refraction is exactly 90 degrees (refracted ray grazes along the surface):

n sin(C) = 1 x sin(90 degrees) = 1

sin(C) = 1/n

Equivalently, n = 1/sin(C) (or n = cosec C).

…

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CBSE 2025Set ANNUAL1 mark
$Q. Write the reason for the shining of air bubbles in the water.$
›Reveal solutionSolution
Light travelling from the denser medium (water) towards the rarer medium (air bubble) strikes the curved surface at an angle greater than the critical angle and undergoes total internal reflection, making the bubble appear bright/silvery.

Water is optically denser than air. When light travelling inside water falls on the curved surface of an air bubble, it is going from a denser medium (water) to a rarer medium (air).

For total internal reflection to occur, two conditions must be met:

Light must travel from a denser medium to a rarer medium.

The angle of incidence at the interface must be greater than the critical angle $\theta_c$ for the water–air pair (where $\sin\theta_c = 1/n_{water}$ ). …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. If the critical angle for total internal reflection from a medium to vacuum is 30°, then velocity of light in medium is (velocity of light = 3\times10⁸ ms⁻¹).$
(a) 1.5 × 10⁸ ms⁻¹
(b) 0.75 × 10⁸ ms⁻¹
(c) 3 × 10⁸ ms⁻¹
(d) 2 × 10⁸ ms⁻¹.
›Reveal solutionSolution

Using $\sin C = 1/n$ with C = 30° gives n = 2, so the speed of light in the medium is c/2 = 1.5×10⁸ m/s.

For total internal reflection from a medium into vacuum, the critical angle C satisfies

$\sin C = \frac{1}{n}$ where n is the refractive index of the medium (with respect to vacuum).

$\sin 30° = 0.5 = \frac{1}{n} \Rightarrow n = 2$

…

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CBSE 2025Set ANNUAL1 mark
$Q.In the case of refraction, write down the relation between critical angle and refractive index of the denser medium. OR What will be the change in focal length f of a concave mirror when immersed in a liquid of refractive index μ?$
›Reveal solutionSolution

The refractive index of the denser medium (with respect to the rarer medium) is the reciprocal of the sine of the critical angle.

At the critical angle C, light travelling from a denser to a rarer medium refracts exactly along the interface (angle of refraction = 90°). By Snell's law, applied to the ray going from the denser medium (refrac …

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Q.At what angle should a ray of light be incident on the face of a prism of refracting angle $60^\circ$ so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is $1.524$ .

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: Critical Angle & Prism Geometry Approach

Step 1 — Find the critical angle for the prism material

Step 2 — Relate prism angles

Step 3 — Apply Snell’s law at the first face …

Mistake 1: Confusing the angle of incidence at the first face with the angle at the second face

Mistake 2: Forgetting the prism geometry relation between $r_1$ and $r_2$

Mistake 3: Using the wrong Snell’s law sign or direction at the first face

Mistake 4: Not checking if the calculated $i$ is physically possible

Q.At what angle should a ray of light be incident on the face of a prism of refracting angle 60∘60^\circ60∘ so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.5241.5241.524.

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Q.At what angle should a ray of light be incident on the face of a prism of refracting angle $60^\circ$ so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is $1.524$ .

Unlock everything free for 14 days