Refraction of plane waves

Stand a pencil in a glass of water and it looks snapped at the surface. Ray optics in Class 12's earlier chapter tells you the ray "bends" and hands you Snell's law, $n_1\sin i = n_2\sin r$ , as a rule to memorise and apply. It never tells you why light bends at all. Wave optics does — and the proof is almost embarrassingly geometric, needing nothing more than a wavefront, a ruler, and two right triangles that happen to share a hypotenuse. Huygens' construction pictures light not as a ray but as an advancing wavefront: a surface over which the wave has the same phase everywhere, built moment by moment from a swarm of tiny secondary wavelets. When a broad, flat wavefront meets a plane boundary between two media obliquely, one edge of the wavefront reaches the boundary and starts sending wavelets into the new medium before the other edge arrives. Because the wave travels at a different speed on each side of the boundary, the two edges of the wavefront cover different distances in the very same interval of time — and a wavefront that is no longer straight-on to its old direction is, by definition, a wavefront travelling in a new direction. That tilt is refraction. This article rebuilds that construction from scratch, turns the geometry into $\sin i/\sin r = v_1/v_2$ , and reads Snell's law straight off it using $n = c/v$ . Two worked examples carry the same air-into-glass and A-into-B numbers you may already have seen, but now with an extra layer: wavelength change across the boundary and a competitive-style multi-hop refraction. Along the way we fix the single most common exam slip in this topic — mixing up which triangle side belongs to the incident angle and which belongs to the refracted angle — and the equally common myth that a "denser" medium somehow means the wave speeds up.

Key takeaways

Wave optics explains refraction, ray optics only states it. Ray optics gives you $n_1\sin i = n_2\sin r$ ; Huygens' construction shows where that equation comes from — unequal wave speeds on the two sides of a boundary.
The core geometry. A wavefront $AB$ hits boundary $XY$ at incidence angle $i$ ; while $B$ covers $BA' = v_1\tau$ in medium 1, the wavelet from $A$ covers $AA'' = v_2\tau$ in medium 2, both triangles sharing hypotenuse $AA'$ .
The ratio falls straight out. $\sin i/\sin r = v_1/v_2$ , and since $n = c/v$ , this converts directly into Snell's law, $n_1\sin i = n_2\sin r$ .
Speed and refractive index are inverses. A denser medium (higher $n$ ) always means a slower wave ( $v = c/n$ ), never a faster one — this is the single most tested misconception.
Frequency never changes; wavelength does. Crossing a boundary changes speed and wavelength ( $\lambda = v/f$ ) together, but the frequency $f$ , set by the source, stays fixed on both sides.
The bend is always towards the normal when entering a denser medium, and away from the normal when leaving it — both are direct readouts of $v_2 < v_1$ or $v_2 > v_1$ in the same construction.
Key idea: refraction is the wavefront tilting because its two edges travel at different speeds v1 and v2 on either side of a boundary in the same time, giving sin i / sin r = v1/v2, which is Snell's law n1 sin i = n2 sin r once you substitute n = c/v.

The core insight

A plane wavefront refracts — bends at a boundary — because its two edges advance at the two different wave speeds $v_1$ and $v_2$ of the two media in the same time interval, and working out that geometry gives $\sin i/\sin r = v_1/v_2 = n_2/n_1$ , which is exactly Snell's law.

Where it fits in the syllabus

This topic is the second pillar of the CBSE "Wave Optics" chapter, built directly on Huygens' principle and immediately preceding reflection of plane waves in the same construction family. The prescribed syllabus explicitly asks you to prove the laws of refraction (and reflection) of a plane wave at a plane surface using Huygens' wavefronts — this is one of the few full derivations examined in this chapter, unlike the fringe-width and diffraction results later in the same unit, which are quoted without derivation. Note also that polarisation of light (Malus's law, Brewster's angle, polaroids) has been dropped from the 2025-26 syllabus entirely and plays no role here or anywhere in this chapter.

The physical picture

Picture a broad, flat wavefront $AB$ travelling through medium 1 (wave speed $v_1$ ) and striking a plane boundary $XY$ obliquely, making an angle of incidence $i$ with the normal. By Huygens' principle, every point on a wavefront is itself a source of secondary spherical wavelets, and the new wavefront a little later is the common tangent — the envelope — to all those wavelets.

BA' = v_1\tau, \qquad AA'' = v_2\tau

where: $\tau$ is the short time interval during which point $B$ of the original wavefront (still travelling in medium 1) reaches the boundary at a new point $A'$ , and during which the wavelet launched from point $A$ (already sitting on the boundary at the instant $B$ was there) has advanced a distance $AA''$ into medium 2 at that medium's own speed $v_2$ . Physically: the two edges of the same wavefront are racing across the boundary region at two different speeds, because one edge is still in medium 1 while the other has already crossed into medium 2.

\sin i = \frac{BA'}{AA'}, \qquad \sin r = \frac{AA''}{AA'}

where: both right triangles, $\triangle ABA'$ and $\triangle AA''A'$ , share the same hypotenuse $AA'$ — the straight-line path along the boundary connecting the two points where the construction is anchored. The incidence angle $i$ sits inside the first triangle (medium 1's side of the boundary), and the refraction angle $r$ sits inside the second (medium 2's side). This shared hypotenuse is the entire trick of the proof: it lets you divide one equation by the other and have $AA'$ cancel out completely.

Dividing one sine expression by the other, the shared hypotenuse $AA'$ cancels and the geometry collapses to a single ratio of speeds — the wave-optics statement of the law of refraction. Because $v_2 < v_1$ when medium 2 is optically denser, $r < i$ automatically, which is exactly the "bends towards the normal" rule you already use in ray optics, now derived rather than assumed.

Derivation ARefraction of a plane wavefront at a plane boundary

The goal of this derivation is to start purely from Huygens' picture of a wavefront advancing through space at a definite speed in each medium, and to show — using only two right-angled triangles sharing a common hypotenuse and the everyday definition of sine as opposite over hypotenuse — that the ratio $\sin i/\sin r$ must equal the ratio of the two wave speeds $v_1/v_2$ , and that substituting the refractive-index definition $n = c/v$ then converts this purely geometric statement into the familiar algebraic form of Snell's law, $n_1\sin i = n_2\sin r$ , without appealing to rays or particles at any stage.

Step 1. Set up the wavefront and the boundary. Let $AB$ be a plane wavefront in medium 1 (speed $v_1$ ) approaching a plane boundary $XY$ , making angle of incidence $i$ with the normal to $XY$ . At the instant point $A$ of the wavefront just touches the boundary, point $B$ is still a perpendicular distance away, travelling towards the boundary.

Step 2. Let both ends evolve for the same short time $\tau$ . In this interval, $B$ travels through medium 1 and reaches the boundary at a new point $A'$ , covering a distance $BA' = v_1\tau$ . Simultaneously, since $A$ has already reached the boundary, Huygens' principle says $A$ becomes a source of a secondary wavelet spreading into medium 2 at that medium's speed $v_2$ ; by the same instant, this wavelet has radius $AA'' = v_2\tau$ .

Step 3. Build the two right triangles sharing hypotenuse $AA'$ . The refracted wavefront is the tangent from $A'$ to the wavelet centred on $A$ , so the new wavefront is the line $A'A''$ , and $\angle AA''A' = 90°$ . Meanwhile the original wavefront's geometry gives $\angle ABA' = 90°$ . Both triangles $\triangle ABA'$ and $\triangle AA''A'$ have $AA'$ as their common hypotenuse, with $i$ the angle at $A$ in the first triangle and $r$ the angle at $A'$ in the second.

Step 4. Write each sine and divide. From the two right triangles, $\sin i = BA'/AA' = v_1\tau/AA'$ and $\sin r = AA''/AA' = v_2\tau/AA'$ . Dividing the first by the second, both $\tau$ and $AA'$ cancel exactly, leaving $\dfrac{\sin i}{\sin r} = \dfrac{v_1}{v_2}$ — a pure statement about wave speeds, with no reference to any ray.

Step 5. Convert to Snell's law using the refractive index. The refractive index of a medium relative to vacuum is defined as $n = c/v$ , so $v_1 = c/n_1$ and $v_2 = c/n_2$ . Substituting, $\dfrac{\sin i}{\sin r} = \dfrac{c/n_1}{c/n_2} = \dfrac{n_2}{n_1}$ , which rearranges to the familiar form

n_1\sin i = n_2\sin r.

Because $n_2 > n_1$ makes the wave slower in medium 2 ( $v_2 < v_1$ ), the same algebra forces $\sin r < \sin i$ , i.e. the wavefront (and the equivalent ray) bends towards the normal on entering a denser medium — the physical picture and the algebra agree by construction, not by coincidence.

Worked examples

Board level - speed and angle across an air-glass boundary

A plane wavefront in air ( $v_1 = 3\times10^8\ \text{m/s}$ ) is incident on a glass slab of refractive index $n_2 = 1.5$ at an angle of incidence $i = 30°$ . Find (a) the speed of the wave in glass and (b) the angle of refraction $r$ .

Step 1. Find $v_2$ from the definition of refractive index. Since $n_2 = c/v_2$ (with $n_1 = 1$ for air, so $c \approx v_1$ ),

v_2 = \frac{c}{n_2} = \frac{3\times10^8}{1.5} = 2\times10^8\ \text{m/s}.

Step 2. Apply the wave-speed form of Snell's law, $\sin i/\sin r = v_1/v_2$ :

\sin r = \frac{v_2}{v_1}\sin i = \frac{2\times10^8}{3\times10^8}\times \sin 30° = \frac{2}{3}\times 0.5 = 0.333.

Step 3. Invert to get the angle. $r = \sin^{-1}(0.333) \approx 19.5°$ . Since $r < i$ , the wavefront has bent towards the normal, exactly as expected for light entering a denser medium.

Final answers: $v_2 = 2\times10^8\ \text{m/s}$ and $r \approx 19.5°$ .

Competitive level - wavelength change and a two-boundary hop

Yellow sodium light of frequency $f = 5\times10^{14}\ \text{Hz}$ travels from medium A ( $n_A = 1.2$ ) into medium B ( $n_B = 1.6$ ) and then on into medium C ( $n_C = 2.0$ ), each boundary being parallel to the others. Take $c = 3\times10^8\ \text{m/s}$ . Find (a) the wave speed and wavelength in each of the three media, and (b) if the angle of incidence in A is $i = 40°$ , find the angle the wavefront makes with the normal inside medium C directly, without separately computing the angle inside B.

Step 1. Compute the three speeds from $v = c/n$ . $v_A = 3\times10^8/1.2 = 2.5\times10^8\ \text{m/s}$ ; $v_B = 3\times10^8/1.6 = 1.875\times10^8\ \text{m/s}$ ; $v_C = 3\times10^8/2.0 = 1.5\times10^8\ \text{m/s}$ . The wave slows down at every boundary, since $n_A < n_B < n_C$ .

Step 2. Compute the three wavelengths from $\lambda = v/f$ , using the fact that $f$ stays fixed at $5\times10^{14}\ \text{Hz}$ throughout. $\lambda_A = v_A/f = 500\ \text{nm}$ ; $\lambda_B = v_B/f = 375\ \text{nm}$ ; $\lambda_C = v_C/f = 300\ \text{nm}$ . Speed and wavelength both shrink together while frequency never changes.

Step 3. Chain Snell's law across both boundaries directly. Because $n_A\sin i = n_B\sin(\text{angle in B})$ and $n_B\sin(\text{angle in B}) = n_C\sin(\text{angle in C})$ , the middle term is common to both, so it also equals $n_A\sin i = n_C\sin(\text{angle in C})$ — the intermediate medium's angle never needs to be found explicitly for parallel boundaries.

Step 4. Solve for the angle in C. $\sin(\text{angle in C}) = \dfrac{n_A}{n_C}\sin i = \dfrac{1.2}{2.0}\times \sin 40° = 0.6\times 0.643 = 0.386$ , so the angle in C is $\sin^{-1}(0.386) \approx 22.7°$ .

Final answers: $v_A = 2.5\times10^8\ \text{m/s}$ , $v_B = 1.875\times10^8\ \text{m/s}$ , $v_C = 1.5\times10^8\ \text{m/s}$ ; $\lambda_A = 500\ \text{nm}$ , $\lambda_B = 375\ \text{nm}$ , $\lambda_C = 300\ \text{nm}$ ; angle in C $\approx 22.7°$ . Timing note: on a chained multi-boundary problem like this, resist the urge to compute the intermediate angle in B — for parallel boundaries the product $n\sin\theta$ is conserved straight through, so jump directly from the first medium's $n_A\sin i$ to the last medium's $n_C\sin\theta_C$ and save a full extra line of working under exam time pressure.

Common traps

Mixing up which side of which triangle gives $\sin i$ and which gives $\sin r$ — always identify the shared hypotenuse $AA'$ first (the line joining the two points where the construction is anchored on the boundary), then read $i$ off the medium-1 triangle and $r$ off the medium-2 triangle.
Believing a denser medium means a faster wave — it is the reverse: a higher refractive index always means a lower wave speed, since $n = c/v$ puts $n$ in the numerator and $v$ in the denominator.
Forgetting that refractive index compares speeds, not wavelengths or frequencies directly — frequency is fixed by the source and never changes across a boundary, while both speed and wavelength change together, in the same proportion, so $\lambda = v/f$ stays consistent on both sides.
Drawing the incident and refracted wavefronts on the same side of the boundary, or drawing the secondary wavelet's radius pointing the wrong way, instead of showing the wavefront visibly tilting as it crosses from one medium into the other.
On multi-boundary problems, wasting time computing every intermediate angle instead of recognising that $n\sin\theta$ is conserved straight across any stack of parallel boundaries.

You should be able to…

Draw the full Huygens construction for a plane wavefront refracting obliquely at a plane boundary, correctly labelling $A$ , $B$ , $A'$ , $A''$ , $i$ , and $r$ .
Reproduce the five-step derivation from $BA' = v_1\tau$ and $AA'' = v_2\tau$ through to $n_1\sin i = n_2\sin r$ , explaining why the shared hypotenuse is the key step.
Explain, in terms of wave speed alone, why a wavefront bends towards the normal entering a denser medium and away from it leaving one.
Solve for an unknown speed, refractive index, or angle given the other two, using either $\sin i/\sin r = v_1/v_2$ or $n_1\sin i = n_2\sin r$ .
State and justify that frequency is invariant across a refracting boundary while speed and wavelength change together, via $\lambda = v/f$ .
Chain Snell's law across two or more parallel boundaries without computing every intermediate angle.

Ask the tutor: Want to walk through the Huygens construction diagram step by step, or practise chaining Snell's law across three or four media until it is automatic?

Why does the derivation use a wavefront and secondary wavelets instead of just drawing a bent ray like in Class 10?

A ray is only a shorthand for the direction a wavefront is moving — it does not, by itself, explain why the direction changes. The wavefront picture shows the actual mechanism: one edge of the wavefront is still crossing medium 1 while the other edge has already started spreading into medium 2 at a different speed, and it is this speed mismatch across the width of the wavefront that physically tilts it. Once you see the tilt, drawing the equivalent bent ray (always perpendicular to the wavefront) is just a convenient summary, not the explanation itself.

If frequency never changes, why do prisms split white light into colours by wavelength, not frequency?

Good catch — both descriptions are consistent. The refractive index of glass actually depends slightly on frequency (a real-world subtlety this simple construction glosses over, called dispersion), so violet light and red light bend by very slightly different amounts even though each colour's own frequency is unchanged across the boundary. Inside the glass, each colour also has its own wavelength, since $\lambda = v/f$ and $v$ differs slightly per colour. It is these small differences in refractive index, hence in bending angle, that a prism uses to fan white light into a spectrum — a topic that belongs to the earlier Ray Optics chapter, not to this wave-optics proof itself.

Is this the same Snell's law from Class 10, just with a fancier proof, or is something actually new here

The equation $n_1\sin i = n_2\sin r$ is identical to the one you used in Class 10 — nothing about the final formula changes. What is new is the justification: in Class 10 it was handed to you as an experimental rule to apply; here it falls out as a forced consequence of light being a wave that travels at different speeds in different media. This matters for CBSE because the syllabus explicitly lists "prove the law of refraction using Huygens principle" as an examinable derivation, so you are expected to reproduce Steps 1 through 5 above, not merely quote the final equation.

Does the direction of bending ever reverse, say if I go from glass back into air?

Yes, and the same construction covers it automatically — you do not need a separate rule. Going from a denser medium (higher $n$ , lower $v$ ) into a rarer one (lower $n$ , higher $v$ ) simply swaps which speed is larger, so the wavefront now bends away from the normal instead of towards it. Re-run Step 4 of the derivation with $v_1 < v_2$ this time and you will see $\sin r > \sin i$ fall out directly, confirming the reversal algebraically rather than needing to memorise it as a separate case.