Obtain the equation for radius of illumination (or) Snell's window.

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Q. Obtain the equation for radius of illumination (or) Snell's window.

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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 It Matters — Real-World Examples

Optical fibres: A glass core (high $n$ ) surrounded by a cladding (lower $n$ ). Light injected at a steep angle bounces down the fibre via repeated TIR. No light leaks out, even if the fibre is bent. This is how internet data travels across oceans.

Diamonds sparkle: Diamond has a very high refractive index ( $n \approx 2.42$ ), so its critical angle is tiny ( $\approx 24.4^\circ$ ). Light entering a diamond gets trapped inside, bouncing around many times before escaping. That multiple internal reflection creates the brilliant sparkle.

Mirage on a hot road: Hot air near the road has lower $n$ than cooler air above. Light from the sky can undergo TIR at the hot-air layer, creating the illusion of water on the road.

One-Line Summary

Total internal reflection is the complete reflection of light back into a denser medium when it strikes the boundary with a rarer medium at an angle greater than the critical angle.

Total internal reflection and the critical angle condition, sin C = n_rarer/n_denser, are core topics in the NCERT Class 12 Physics chapter on ray optics, tested extensively in CBSE boards, JEE Main and NEET, with optical fibres and diamond sparkle as classic real-world examples. Anyone searching "total internal reflection definition critical angle formula class 12 physics" will find this denser-to-rarer explanation matches the NCERT-prescribed treatment exactly.

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 ( $n_1 > n_2$ )
Angle of incidence exceeds the critical angle ( $\theta_1 > \theta_c$ )

Where:

\theta_c = \sin^{-1}\left(\frac{n_2}{n_1}\right)

5. Why This Matters (Conceptual Insight)

Think of it like a runner trying to jump from a fast surface (denser medium) onto a slow surface (rarer medium). At shallow angles, they can "refract" (change direction). But at steep angles, they simply cannot enter the slower medium — they bounce back.

The formula $\sin \theta_c = n_2/n_1$ is the mathematical expression of this physical limit.

Quick Exam Tip

Scenario	Formula	Why
Critical angle	$\sin \theta_c = \frac{n_2}{n_1}$	Snell's Law with $\theta_2 = 90^\circ$
TIR condition	$\theta_1 > \theta_c$ and $n_1 > n_2$	Beyond Snell's Law's physical limit

Never memorise blindly — the critical angle formula is just Snell's Law at the extreme case. Derive it if you forget!

Step 1. A point (source or eye) sits at depth $d$ below a water surface (refractive index $n_1=n$ , outer medium air, $n_2=1$ ). Light reaching the surface at exactly the critical angle $i_c$ grazes along the boundary ( $r=90°$ ); Snell's law in product form gives $n_1\sin i_c=n_2\sin90°=1$ , i.e. $\sin i_c=1/n$ .

Step 2. From the right triangle formed by the depth $d$ (vertical leg), the radius $R$ of the illuminated circle (horizontal leg) and the slant distance to the edge of the circle (hypotenuse), $\sin i_c=\dfrac{R}{\sqrt{d^2+R^2}}$ .

Step 3. Equating the two expressions for $\sin i_c$ : $\dfrac{1}{n}=\dfrac{R}{\sqrt{d^2+R^2}}$ . Squaring both sides: $\dfrac{1}{n^2}=\dfrac{R^2}{d^2+R^2}$ , so $d^2+R^2=n^2R^2$ , giving $d^2=R^2(n^2-1)$ .

Step 4. Solving for $R$ : $\boxed{R=\dfrac{d}{\sqrt{n^2-1}}}$ -- equivalently, since $\tan i_c=\sin i_c/\cos i_c$ and $\cos i_c=\sqrt{1-1/n^2}=\sqrt{n^2-1}/n$ , this is the same as $R=d\tan i_c$ .

✓Final answer

R=\dfrac{d}{\sqrt{n^2-1}}

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Q.Obtain the equation for radius of illumination (or) Snell's window.

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