Snell's Law and Total Internal Reflection

What Snell’s Law Tells Us About Light

When light moves from one medium to another — say, from air to water, or from glass to air — it bends. The amount of bending follows a precise rule discovered by Willebrord Snell in 1621. We use this rule to design lenses, fibre optics, prisms and even diamond cuts.

Snell’s law and its dramatic offshoot, total internal reflection (TIR), sit at the heart of Class 10 and Class 12 optics. They show up in CBSE boards, NEET, JEE Main, and even Olympiad-style problems. Mastering them is the cleanest path to scoring on ray optics.

We’ll cover the law, the geometry, the conditions for TIR, and the formula bag every JEE/NEET aspirant needs.

Key Terms & Definitions

Refractive index ( $n$ ) — the ratio of speed of light in vacuum to the speed of light in the medium. Vacuum: $n = 1$ . Water: $n \approx 1.33$ . Glass: $n \approx 1.5$ . Diamond: $n \approx 2.42$ .

Angle of incidence ( $\theta_i$ ) — the angle between the incident ray and the normal to the surface at the point of incidence.

Angle of refraction ( $\theta_r$ ) — the angle between the refracted ray and the normal in the second medium.

Critical angle ( $\theta_c$ ) — the angle of incidence (in the denser medium) for which the angle of refraction equals $90°$ . Beyond this, no refraction happens; all light reflects back.

Optical density vs mass density — careful, these are different. Optical density refers to refractive index; a denser medium optically need not be denser by mass.

Methods/Concepts

Snell’s Law — the basic statement

When light passes from medium 1 (refractive index $n_1$ ) to medium 2 (refractive index $n_2$ ):

n_1 \sin\theta_1 = n_2 \sin\theta_2

The light bends towards the normal when entering a denser medium ( $n_2 > n_1$ ), and away from the normal when entering a rarer medium.

Why does light bend?

Light travels slower in denser media. When a wavefront enters a denser medium at an angle, the side that hits first slows down first, while the other side keeps moving faster. This skews the wavefront direction — that’s the bending.

This wavefront argument was Huygens’s, and it gives Snell’s law as a direct consequence. Fermat’s principle (light takes the path of least time) gives the same result.

Critical Angle and TIR

When light goes from a denser medium to a rarer medium, it bends away from the normal. As $\theta_1$ increases, $\theta_2$ increases faster — until $\theta_2$ hits $90°$ . The corresponding $\theta_1$ is the critical angle:

\sin\theta_c = \frac{n_2}{n_1}

(with $n_1 > n_2$ , i.e., light starts in the denser medium.)

For $\theta_1 > \theta_c$ , refraction is impossible. All the light bounces back into the denser medium — that is total internal reflection.

Conditions for TIR

Two conditions, both required:

Light must travel from denser to rarer medium.
The angle of incidence must exceed the critical angle.

If either fails, you get partial refraction + partial reflection (the usual case), not TIR.

Solved Example 1 — Basic Snell’s law (Easy, CBSE)

A ray of light enters water from air at $30°$ from the normal. Find the angle of refraction. ( $n_{\text{water}} = 1.33$ .)

$n_1 = 1$ (air), $n_2 = 1.33$ (water), $\theta_1 = 30°$ .

$1 \times \sin 30° = 1.33 \times \sin\theta_2$ .

$\sin\theta_2 = 0.5 / 1.33 \approx 0.376$ .

$\theta_2 = \arcsin(0.376) \approx 22.1°$ .

The ray bends towards the normal — water is denser, so this makes sense.

Solved Example 2 — Critical angle (Medium, JEE Main)

Find the critical angle for light going from glass ( $n = 1.5$ ) to air.

$\sin\theta_c = n_{\text{air}} / n_{\text{glass}} = 1/1.5 = 0.667$ .

$\theta_c = \arcsin(0.667) \approx 41.8°$ .

For any $\theta_i > 41.8°$ in the glass, light cannot exit — it reflects back. This is the principle behind glass prisms used in periscopes.

Solved Example 3 — Optical fibre (Hard, JEE Advanced)

An optical fibre has a core of refractive index $n_1 = 1.5$ and cladding of $n_2 = 1.45$ . Find the maximum angle of acceptance — the largest angle at which a ray entering the front face from air gets totally internally reflected inside.

$\sin\theta_c = n_2/n_1 = 1.45/1.5 = 0.967$ . So $\theta_c \approx 75.2°$ .

For TIR, the ray must hit the core-cladding boundary at $\geq \theta_c$ from the normal — that means it must travel at $\leq 90° - \theta_c \approx 14.8°$ from the fibre’s axis.

Air to core: $1 \times \sin\theta_{\max} = 1.5 \times \sin(14.8°)$ .

$\sin\theta_{\max} = 1.5 \times 0.255 = 0.383$ .

$\theta_{\max} \approx 22.5°$ .

This is the half-angle of the acceptance cone — a key spec for fibre-optic communications.

Exam-Specific Tips

JEE Main: Snell’s law and TIR appear in roughly 1 question per paper. Common patterns: prism deviation, glass slab apparent depth, optical fibre acceptance angle.

NEET: Snell’s law numericals are nearly guaranteed. Critical angle questions show up at least once every two years. Memorise critical angles for water-air ( $\approx 48.8°$ ), glass-air ( $\approx 41.8°$ ), diamond-air ( $\approx 24.4°$ ).

CBSE Class 10: Focus on the qualitative behaviour and one numerical. CBSE Class 12 expects you to derive critical angle and explain TIR in fibres and prisms.

Common Mistakes to Avoid

Mistake 1: Measuring angles from the surface instead of the normal. Snell’s law uses angles from the normal (perpendicular to the surface). If the question gives the angle from the surface, subtract from $90°$ first.

Mistake 2: Applying TIR when going from rare to dense. TIR only happens going from dense to rare. Swap the formula direction here and your answer is nonsense.

Mistake 3: Forgetting $n_1 \sin\theta_1 = n_2 \sin\theta_2$ requires both sides of the boundary in your equation. Students sometimes write $\sin\theta_1 = n_2 \sin\theta_2$ (dropping $n_1$ ). Always include both refractive indices.

Mistake 4: Using $\tan\theta$ instead of $\sin\theta$ . The geometry is built on $\sin$ , not $\tan$ . (You may be confusing with Brewster’s angle, where $\tan$ shows up.)

Mistake 5: Forgetting that critical angle only depends on the ratio $n_2/n_1$ . Doubling both refractive indices doesn’t change $\theta_c$ .

Practice Questions

Q1. Light enters glass ( $n = 1.5$ ) from air at $60°$ . Find the angle of refraction.

$\sin 60° = 1.5 \sin\theta_r \Rightarrow \sin\theta_r = 0.866/1.5 = 0.577 \Rightarrow \theta_r = 35.3°$ .

Q2. A diver looks up at the sky from underwater. Beyond what angle from the vertical does the surface appear silvered?

Critical angle from water to air: $\sin\theta_c = 1/1.33 \Rightarrow \theta_c \approx 48.8°$ . Beyond this, the surface acts like a mirror by TIR.

Q3. A glass prism has refractive index $1.5$ . What is the critical angle at the glass-air interface?

$\sin\theta_c = 1/1.5 \approx 0.667 \Rightarrow \theta_c \approx 41.8°$ .

Q4. Light goes from water ( $n = 1.33$ ) to glass ( $n = 1.5$ ) at an angle of $45°$ . Find the angle of refraction.

$1.33 \sin 45° = 1.5 \sin\theta_r \Rightarrow \sin\theta_r = 1.33 \times 0.707/1.5 = 0.627 \Rightarrow \theta_r = 38.8°$ .

Q5. A swimming pool appears $1.5$ m deep when viewed from straight above. If the actual depth is $d$ and water has $n = 1.33$ , find $d$ .

Apparent depth = real depth / $n$ . So $1.5 = d/1.33 \Rightarrow d \approx 2$ m.

Q6. Why do diamonds sparkle so much?

Diamond has refractive index $\approx 2.42$ , so its critical angle is only $\approx 24.4°$ . Light entering the diamond undergoes TIR many times inside, exiting with high concentration through the cuts. Hence the sparkle.

Q7. A ray hits a glass-air boundary at $50°$ . Glass has $n = 1.5$ . Does the ray exit, and if not, what happens?

Critical angle for glass-air is $41.8°$ . Since $50° > 41.8°$ , the ray undergoes TIR and stays inside the glass.

Q8. Find the critical angle between two media with $n_1 = 1.6$ and $n_2 = 1.2$ .

$\sin\theta_c = 1.2/1.6 = 0.75 \Rightarrow \theta_c \approx 48.6°$ .

FAQs

Q: Does Snell’s law work for all wavelengths?

A: Yes, but $n$ depends on wavelength (this is dispersion). Red light bends less than blue light through a prism because $n_{\text{red}} < n_{\text{blue}}$ .

Q: Can light escape during TIR if there is a thin gap of air on the other side?

A: Counter-intuitively, yes — a small fraction of light tunnels through (frustrated TIR). The intensity decays exponentially with the gap thickness. Used in some sensors and fibre couplers.

Q: Is TIR perfect — does no energy leak out?

A: For ideal interfaces in homogeneous media, TIR is theoretically lossless. Real fibres lose energy through impurities and microbends, but TIR itself is loss-free.

Q: How does the critical angle change with wavelength?

A: Since $n$ depends on $\lambda$ , so does $\theta_c$ . For glass, longer wavelengths have slightly lower $n$ and slightly larger $\theta_c$ .

Q: Why does Snell’s law fail at very large angles in light scattering experiments?

A: It doesn’t fail — it just predicts complex angles for $\sin\theta > 1$ , which physically corresponds to TIR. The formula self-consistently signals “no refracted ray exists”.

Q: Are mirages explained by Snell’s law?

A: Yes. Hot air near the road has lower $n$ than cooler air above. Light from the sky bends as it crosses these layers, eventually undergoing TIR — that’s the “water on the road” you see.

Q: How do optical fibres exploit TIR?

A: The core has higher $n$ than the cladding. Light travelling at small angles to the fibre axis hits the core-cladding boundary at $\theta > \theta_c$ , undergoing TIR repeatedly down the length of the fibre.

Q: Is there a critical angle for light going from rare to dense?

A: No. From rare to dense, the refraction angle is always smaller than the incidence angle, so refraction always happens. TIR requires dense-to-rare.