A small bulb is placed at the bottom of a tank containing water to a depth of $80\ \text{cm}$. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is $1.33$. (Consider the bulb to be a point source.)

A small bulb is placed at the bottom of a tank containing water to a depth of 80cm. What is the area of the su

Exercises · 9.5

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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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Light from a point source on the tank floor can only escape through a circular patch directly above it - bounded by the critical angle for the water-air interface. For a depth of $80\ \text{cm}$ and $n=1.33$ , this circle has an area of about $2.6\times10^4\ \text{cm}^2$ ( $\approx2.6\ \text{m}^2$ ).

Why only a circular patch lets light out

Light travelling from water (denser, $n=1.33$ ) to air (rarer, $n=1$ ) bends away from the normal. Beyond a certain critical angle $i_c$ , the refracted ray would have to bend more than $90^\circ$ from the normal - which is impossible - so instead the light undergoes total internal reflection and never leaves the water. Only rays that strike the surface at angles up to $i_c$ actually emerge.

From a point source at the bottom, rays spread out in every direction; the ones that manage to escape trace out a cone (apex at the bulb, half-angle $i_c$ ) whose base is a circle on the water's surface, directly above the source.

Step 1: find the critical angle

\sin i_c = \frac{n_{\text{air}}}{n_{\text{water}}} = \frac{1}{1.33} \approx 0.7519 \quad\Longrightarrow\quad i_c \approx 48.75^\circ.

Step 2: relate the radius of the circle to the depth

The ray that just grazes the critical angle traces the edge of the escaping cone. In the right triangle formed by the bulb, the point directly above it, and the edge of the circle on the surface:

\tan i_c = \frac{r}{h}, \qquad h = 80\ \text{cm}.

\tan i_c = \frac{\sin i_c}{\cos i_c} = \frac{0.7519}{\sqrt{1-0.7519^2}} = \frac{0.7519}{0.6593} \approx 1.140.

r = h\tan i_c = 80\times1.140 \approx 91.2\ \text{cm}.

Step 3: compute the area …

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Common Mistakes & How to Avoid Them

Mistake 1: Confusing Real Depth with Apparent Depth

The error: Students often treat the bulb's actual depth ( $80\ \text{cm}$ ) as the object distance for the refraction formula directly, without considering that the image formed by refraction is virtual and at a different location.

Why it's wrong: For a point source at the bottom, the rays emerging into air appear to come from a virtual image above the actual bulb. The critical angle condition depends on the real depth, not the apparent depth.

How to avoid: Always draw the ray diagram. The bulb is at real depth $h = 80\ \text{cm}$ . The critical angle $\theta_c$ is determined by Snell's law at the water-air interface:

\sin\theta_c = \frac{1}{n} = \frac{1}{1.33}

The radius $r$ of the circular patch on the water surface is:

r = h \tan\theta_c

Key: Use real depth $h$ , not apparent depth.

Mistake 2: Using $\sin\theta_c = n_2/n_1$ Incorrectly

The error: Writing $\sin\theta_c = \frac{n_{\text{water}}}{n_{\text{air}}}$ instead of $\frac{n_{\text{air}}}{n_{\text{water}}}$ .

Why it's wrong: For total internal reflection, light travels from denser (water) to rarer (air) medium. The critical angle formula is:

\sin\theta_c = \frac{n_{\text{rarer}}}{n_{\text{denser}}} = \frac{1}{1.33}

How to avoid: Always identify which medium light is leaving (denser) and which it is entering (rarer). The smaller refractive index goes in the numerator.

Mistake 3: Forgetting the Circular Geometry

The error: After finding $\theta_c$ , students sometimes use $r = h \sin\theta_c$ or $r = h / \tan\theta_c$ .

Why it's wrong: From the geometry (right triangle with height $h$ and base $r$ ):

\tan\theta_c = \frac{r}{h} \quad \Rightarrow \quad r = h \tan\theta_c

How to avoid: Draw the triangle: vertical side = depth $h$ , horizontal side = radius $r$ , angle at the bulb = $\theta_c$ . Then apply $\tan$ .

Mistake 4: Calculating Area Incorrectly

The error: Using $A = \pi r$ or $A = 2\pi r$ instead of $A = \pi r^2$ . …

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

AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
$\sqrt{3}$
›Reveal solutionSolution
This is the classic right-angle-prism (corner-reflector) geometry where the ray strikes each leg at $45^\circ$ ; total internal reflection at that angle requires $n\ge\sqrt2$ .

Concept and Intuition

In a right-angled isosceles prism with the right angle at $Q$ and $PQ=QR$ , a ray entering through the hypotenuse $PR$ and reflecting off both legs $PQ$ and $QR$ (as shown, reflecting first off $PQ$ then off $QR$ before exiting through $PR$ ) does so at exactly $45^\circ$ to the normal at each leg — this is simple geometry of the isosceles right triangle, independent of where exactly on the hypotenuse the ray enters, as long as it undergoes this two-bounce path (this is precisely the working principle of the 'porro prism' corner reflectors used in binoculars and periscopes). For the reflections to be total internal reflection (not partial, lossy reflection), the angle of incidence at each leg ( $45^\circ$ ) must be at least the critical angle $\theta_c$ of the glass.

Step-by-Step Solution

Geometry: since $PQ\perp QR$ and the ray path is symmetric (reflecting off $PQ$ then $QR$ , and by symmetry of the isosceles right triangle with $PQ=QR$ ), the angle of incidence at each leg works out to $45^\circ$ from the normal.

For total internal reflection to actually occur (rather than the light partially escaping through the legs), we need the angle of incidence to be at least the critical angle: $45^\circ \ge \theta_c$ .

The critical angle relates to refractive index via $\sin\theta_c = \dfrac1n$ . …
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AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
$Q. The following statement is true in case of total internal reflection? (A) Light must travel from rarer medium to denser medium & angle of incidence should be greater than critical angle (B) Light must travel from denser medium to rarer medium & angle of incidence should be less than critical angle (C) Light must travel from denser medium to rarer medium & angle of incidence should be > 90° (D) Light must travel from denser medium to rarer medium & angle of incidence should be greater than critical angle$
›Reveal solutionSolution
Total internal reflection requires travel from denser to rarer medium with the angle of incidence exceeding the critical angle — exactly what option (D) states.

Concept and Intuition

At an interface between a denser (higher refractive index) and rarer (lower index) medium, Snell's law $n_1\sin\theta_1=n_2\sin\theta_2$ means the refracted ray bends away from the normal. As the incidence angle increases, the refraction angle reaches 90° at the critical angle $\theta_c$ (where $\sin\theta_c=n_2/n_1$ ). Beyond that angle no refracted ray can exist (since $\sin\theta_2$ would have to exceed 1), so all the light reflects back — total internal reflection. This can only happen going from denser to rarer, and only strictly beyond the critical angle (not merely close to 90°).

Step-by-Step Solution

Direction check: TIR needs denser → rarer (light going the other way just refracts further toward the normal, never fully reflects). This eliminates option (A), which has the direction backwards.

Angle check: the condition is angle of incidence $>$ critical angle, not $<$ critical angle — this eliminates option (B). …
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AP EAPCET 2021Set ap-2021-09-06-AN1 markMCQ
$Q.The principle used in the transmission of signals through an optical fibre is (A) Refraction (B) Dispersion (C) Total internal reflection (D) Interference$
›Reveal solutionSolution
Optical fibres transmit light signals by repeated total internal reflection at the core-cladding interface.

Concept and Intuition

An optical fibre has a denser core surrounded by a rarer cladding (lower refractive index). Light entering the fibre within the acceptance angle always strikes the core-cladding boundary at an angle exceeding the critical angle for that interface. Because the angle of incidence exceeds the critical angle, all the light is reflected back into the core (none refracts out) — this is total internal reflection. The process repeats down the entire length of the fibre, letting the signal travel with very low loss even around bends.

Step-by-Step Solution

Recall the conditions for total internal reflection: light must travel from a denser to a rarer medium, and the angle of incidence must exceed the critical angle.

In a fibre, the core (denser) and cladding (rarer) are engineered so that guided rays always meet this condition.

Refraction alone would let light leak out through the cladding — not usable for long-distance transmission. …
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AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
$Q. A well-cut diamond appears bright because ________ (A) it emits light (B) it is radioactive (C) of its total internal reflection (D) it has high density$
›Reveal solutionSolution
A diamond's brilliance comes from total internal reflection at its many facets, enabled by its very high refractive index (small critical angle).

Concept and Intuition

Total internal reflection (TIR) occurs when light travelling in a denser medium strikes the boundary with a less dense medium at an angle greater than the critical angle — instead of refracting out, it reflects entirely back inside. Diamond has an unusually high refractive index (~2.42), giving it a very small critical angle (~24.4°). A skilled cutter angles the diamond's many facets so that light entering from the top undergoes multiple total internal reflections off the bottom facets before finally exiting back out the top, concentrating and scattering the light as sparkle.

Step-by-Step Solution

Diamond does not emit its own light — it only reflects/refracts incoming light, ruling out option (A).

Diamond's brilliance has nothing to do with radioactivity — ruling out option (B).

High density alone (option D) doesn't explain optical brilliance; many dense materials are dull. …
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←Previous question9.4 — Figures 9.27(a) and (b) show refraction of a ray in air incident at 60…→Next question9.6 — A prism is made of glass of unknown refractive index. A parallel beam …

Q.A small bulb is placed at the bottom of a tank containing water to a depth of $80\ \text{cm}$ . What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is $1.33$ . (Consider the bulb to be a point source.)

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)

Why only a circular patch lets light out

Step 1: find the critical angle

Step 2: relate the radius of the circle to the depth

Step 3: compute the area …

Method: Critical Angle & Cone of Emergence

Steps

Common Mistakes & How to Avoid Them

Mistake 1: Confusing Real Depth with Apparent Depth

Mistake 2: Using $\sin\theta_c = n_2/n_1$ Incorrectly

Mistake 3: Forgetting the Circular Geometry

Mistake 4: Calculating Area Incorrectly

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm80\ \text{cm}80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.331.331.33. (Consider the bulb to be a point source.)

Why only a circular patch lets light out

Step 1: find the critical angle

Step 2: relate the radius of the circle to the depth

Step 3: compute the area …

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Q.A small bulb is placed at the bottom of a tank containing water to a depth of $80\ \text{cm}$ . What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is $1.33$ . (Consider the bulb to be a point source.)

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