A square block of glass ABCD has refractive index $1.6$. The corners are arranged with A at the top-left, B at the top-right, C at the bottom-right and D at the bottom-left, so that AB is the horizontal top face, AD is the vertical left face, and the block is viewed edge-on as a square. A pin is embedded at the midpoint of the top face AB. An observer's eye looks into the block through the left face AD. Taking account of refraction and the possibility of total internal reflection at the face AD (for which the critical angle corresponds to refractive index $1.6$), where will the pin appear to be? (a) Appear to be near A (b) Appear to be near D (c) Appear to be at the centre of AD (d) Not be seen at all

A square block of glass ABCD has refractive index 1.6. The corners are arranged with A at the top-left, B at t

NCERT Exemplar · Q13

1.6

(a) Appear to be near A

(b) Appear to be near D

(d) Not be seen at all

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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 the pin can leave through face AD only where it strikes that face at less than the critical angle ( $\approx 38.7^\circ$ for $\mu=1.6$ ). Working through the geometry, those escaping rays all meet AD in its upper portion, near corner A, so the pin is visible and appears near A.

Concept: critical angle and total internal reflection

Going from glass to air, a ray escapes only if its angle of incidence at the surface is less than the critical angle

\theta_c = \sin^{-1}\!\left(\frac{1}{\mu}\right) = \sin^{-1}\!\left(\frac{1}{1.6}\right) = \sin^{-1}(0.625) \approx 38.7^\circ .

Rays hitting AD at more than $\theta_c$ are totally internally reflected and do not reach the observer.

Geometry

Let the square have side $a$ with $D=(0,0)$ , $C=(a,0)$ , $B=(a,a)$ , $A=(0,a)$ . The pin sits at the midpoint of AB, $M=(a/2,\,a)$ . Face AD is the line $x=0$ ; its outward normal is horizontal. A ray from $M$ to a point $P=(0,y)$ on AD makes an angle $\theta$ with that horizontal normal, where

\tan\theta = \frac{a-y}{a/2}.

The ray escapes if $\theta<\theta_c$ , i.e. $\tan\theta < \tan 38.7^\circ \approx 0.80$ :

\frac{a-y}{a/2} < 0.80 \;\Rightarrow\; a-y < 0.40\,a \;\Rightarrow\; y > 0.60\,a.

Result …

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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. …
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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 …
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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:

$\theta = 90^\circ - 60^\circ = 30^\circ$ …
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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 $A_2/A_1$
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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. …
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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: critical angle and total internal reflection

Geometry

Result …

Method: Finding the Escape Window for Light Through a Face (Critical-Angle Geometry)

Steps

Concept and Intuition

Step-by-Step Reasoning

Concept: critical angle and total internal reflection

Geometry

Result …

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