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

80\ \text{cm}

Telangana TsbieTextbookSubjective· 3mImportance★★★★★

18% · 13/73 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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 ( $n_1 > n_2$ ) …

Start your 14-day free trial to unlock the full solution →

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 …

Start your 14-day free trial to unlock the full solution →

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 $A = \pi r^2$ . …

Start your 14-day free trial to unlock the full solution →

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

TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
$48^\circ$
›Reveal solutionSolution
The ray undergoes total internal reflection because the incident angle exceeds the critical angle. The deviation is $84^\circ$ , so option (A) is correct.

The key here is to first check whether the ray refracts into air or reflects back into the medium. The refractive index given is $1.414$ , which is $\sqrt{2}$ — a familiar value. When light travels from a denser medium to a rarer one (air, $n=1$ ), total internal reflection occurs if the angle of incidence exceeds the critical angle. The critical angle $i_c$ is found from Snell’s law at the limiting case where the angle of refraction is $90^\circ$ .

Let’s work through it.

Find the critical angle. Snell’s law: $n_1 \sin i_c = n_2 \sin 90^\circ$ . Here $n_1 = 1.414$ , $n_2 = 1$ . So

$1.414 \sin i_c = 1 \times 1$ $\sin i_c = \frac{1}{1.414} = \frac{1}{\sqrt{2}} \approx 0.7071$
Hence $i_c = 45^\circ$ .

Compare the given angle of incidence with the critical angle.

The incident angle is $48^\circ$ , which is greater than $45^\circ$ . Therefore, the ray does not enter air — it undergoes total internal reflection at the boundary.

Determine the deviation in total internal reflection.

For reflection, the angle of reflection equals the angle of incidence. The ray is turned back into the same medium. The deviation $\delta$ is the angle through which the ray's direction is bent from its original path. …
Start your 14-day free trial to see the full solution →
TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
$120^\circ$
›Reveal solutionSolution
The angle of contact is a material property of the liquid–solid–vapor system and does not depend on the orientation of the solid surface. Therefore, the angle remains $120^\circ$ when the rod is placed horizontally. The correct option is (D).

Concept and Intuition

The angle of contact (or contact angle) is defined as the angle between the tangent to the liquid surface and the solid surface, measured inside the liquid. It is determined by the balance of three interfacial tensions: solid–liquid ( $\gamma_{SL}$ ), solid–vapor ( $\gamma_{SV}$ ), and liquid–vapor ( $\gamma_{LV}$ ). This balance is given by Young’s equation:
$\gamma_{SV} - \gamma_{SL} = \gamma_{LV} \cos\theta$
Here, $\theta$ is the contact angle. Crucially, $\gamma_{SL}$ , $\gamma_{SV}$ , and $\gamma_{LV}$ are intrinsic properties of the materials involved — they depend only on the chemical nature of the solid, the liquid, and the vapor (or air), not on the shape or orientation of the solid.

A common pitfall is to think that tilting the rod changes the “effective” contact angle, but the contact angle is measured locally at the three-phase line (where solid, liquid, and vapor meet). That local geometry is unchanged because the solid’s surface chemistry is the same. The rod’s orientation only changes the macroscopic shape of the meniscus, not the microscopic balance of forces at the contact line.

Step-by-Step Reasoning

Identify the governing principle The contact angle $\theta$ is given by Young’s equation:

$\cos\theta = \frac{\gamma_{SV} - \gamma_{SL}}{\gamma_{LV}}$
The right-hand side depends only on the three interfacial tensions. These tensions are fixed for a given solid, liquid, and surrounding vapor.

Check if orientation changes the interfacial tensions

When the rod is placed vertically, the solid surface is vertical. When placed horizontally, the solid surface is horizontal (or at a different orientation relative to gravity). However, the chemical composition of the rod’s surface does not change — it is the same material. Therefore, $\gamma_{SV}$ , $\gamma_{SL}$ , and $\gamma_{LV}$ remain exactly the same.

Apply Young’s equation …
Start your 14-day free trial to see the full solution →
TG EAPCET 2022Set ap-2022-07-31-FN1 markMCQ
$60^\circ$
›Reveal solutionSolution

The ray meets the side wall at $30^\circ$ to its normal; setting this equal to the critical angle gives $n=\dfrac{1}{\sin30^\circ}=2$ .

Inside the rod the ray travels at $60^\circ$ to the axis. The cylindrical side (rod–air interface) is parallel to the axis, so its normal is perpendicular to the axis. The angle of incidence at that wall, measured from the normal, is:

$\theta = 90^\circ - 60^\circ = 30^\circ$ …
Start your 14-day free trial to see the full solution →
TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
$A_1$
›Reveal solutionSolution
The ratio of maximum to minimum intensity in wave superposition is given by $\left(\frac{A_1+A_2}{A_1-A_2}\right)^2$ . Setting this equal to $9/4$ and solving for $A_2/A_1$ (with $A_1 > A_2$ ) gives $0.20$ , which corresponds to option (B).

Concept & Intuition

When two waves of amplitudes $A_1$ and $A_2$ superimpose, the resultant intensity depends on the phase difference between them. The maximum intensity occurs when the waves are in phase (constructive interference), giving amplitude $A_1 + A_2$ . The minimum intensity occurs when they are exactly out of phase (destructive interference), giving amplitude $|A_1 - A_2|$ . Since intensity is proportional to the square of the amplitude, the ratio of maximum to minimum intensity is $\left(\frac{A_1+A_2}{A_1-A_2}\right)^2$ . The problem gives this ratio as $9:4$ , so we can set up an equation and solve for the ratio of amplitudes.

Step-by-step solution

Write the intensity ratio in terms of amplitudes Maximum intensity $I_{\text{max}} \propto (A_1 + A_2)^2$ Minimum intensity $I_{\text{min}} \propto (A_1 - A_2)^2$ Hence

$\frac{I_{\text{max}}}{I_{\text{min}}} = \left(\frac{A_1 + A_2}{A_1 - A_2}\right)^2$

Set the given ratio The problem states $I_{\text{max}} : I_{\text{min}} = 9 : 4$ , so

$\left(\frac{A_1 + A_2}{A_1 - A_2}\right)^2 = \frac{9}{4}$

Take the square root (positive root, since amplitudes are positive)

$\frac{A_1 + A_2}{A_1 - A_2} = \frac{3}{2}$

Cross-multiply and solve for $A_2/A_1$
Start your 14-day free trial to see the full solution →
TG EAPCET 2021Set ap-2021-08-09-FN1 markMCQ
$Q. Conditions for total internal reflection to occur are:$
a) The ray should travel from rarer to denser medium
b) The ray should travel from denser to rarer medium
c) The angle of incidence should be greater than the critical angle
d) The angle of incidence should be less than the critical angle Which of the following is correct? (A) (b, c) (B) (a, c) (C) (a, d) (D) (b, d)
›Reveal solutionSolution
Total internal reflection requires light to travel from a denser to a rarer medium and the angle of incidence to exceed the critical angle. The correct pair is (b, c), so option (A).

Concept & Intuition

Total internal reflection (TIR) is a phenomenon where light is completely reflected back into the original medium, with no transmission into the second medium. This happens only when light tries to leave a denser medium (like glass or water) into a rarer one (like air). If the angle of incidence is too shallow (less than the critical angle), some light escapes by refraction. But if the angle is steep enough (greater than the critical angle), the refracted ray would have to bend so much that it never actually leaves — instead, all the energy bounces back. So two conditions are necessary: the direction must be from denser to rarer, and the angle must be larger than a specific threshold.

Step-by-step reasoning

Identify the medium condition

TIR cannot occur when light goes from a rarer to a denser medium (e.g., air to glass) because the refracted ray always bends toward the normal, so some light always enters the denser medium. The ray must travel from a denser to a rarer medium.

→ Statement (b) is correct; statement (a) is incorrect.

Identify the angle condition

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°. If the angle of incidence is less than the critical angle, refraction occurs (some light escapes). If it is greater than the critical angle, no refraction is possible — total internal reflection occurs. …
Start your 14-day free trial to see the full solution →

←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 Reasoning

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 …

🎓Unlock everything free for 14 days

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.)

Unlock everything free for 14 days