Huygens principle — wavefront propagation, reflection, refraction derivation

Question

Using Huygens’ principle, derive Snell’s law of refraction. A plane wavefront in medium 1 (speed $v_1$ ) hits a flat boundary with medium 2 (speed $v_2$ ) at angle of incidence $i$ . Show that $\dfrac{\sin i}{\sin r} = \dfrac{v_1}{v_2} = n_{21}$ .

(CBSE 12 Board 5-mark + JEE Main theory)

Solution — Step by Step

Every point on a wavefront acts as a source of secondary wavelets. The new wavefront after time $t$ is the forward envelope (tangent surface) of all these secondary wavelets.

This principle lets us construct the refracted wavefront geometrically.

A plane wavefront AB hits the boundary. Point A reaches the surface first. While the wavelet from A travels into medium 2 a distance $v_2 t$ , point B still needs to travel in medium 1 a distance $v_1 t$ to reach the surface at point C.

In time $t$ : wavelet from A in medium 2 covers $AE = v_2 t$ . The incident wavefront travels from B to C in medium 1: $BC = v_1 t$ .

In triangle ABC (incident side):

\sin i = \frac{BC}{AC} = \frac{v_1 t}{AC}

In triangle AEC (refracted side):

\sin r = \frac{AE}{AC} = \frac{v_2 t}{AC}

Dividing:

\frac{\sin i}{\sin r} = \frac{v_1 t / AC}{v_2 t / AC} = \frac{v_1}{v_2}

Since refractive index $n = c/v$ , we get $n_1 \sin i = n_2 \sin r$ , which is Snell’s law.

flowchart TD
    A["Plane wavefront approaches boundary"] --> B["Point A hits surface first"]
    B --> C["A emits secondary wavelet in medium 2"]
    B --> D["Point B still travelling in medium 1"]
    C --> E["Wavelet from A: radius = v₂t"]
    D --> F["B reaches surface at C: distance = v₁t"]
    E --> G["Draw tangent from C to wavelet circle"]
    G --> H["New refracted wavefront CE formed"]
    H --> I["sin i / sin r = v₁/v₂ = n₂/n₁"]

Why This Works

Huygens’ construction is powerful because it makes no assumptions about the nature of light — it works for any wave. The key insight is that the wavefront bends at a boundary because different parts of it enter the second medium at different times. The part that enters first slows down (or speeds up), while the rest is still travelling at the original speed. This differential speed is what causes the change in direction.

When light goes from a faster medium to a slower medium ( $v_2 < v_1$ ), the refracted wavefront tilts toward the normal ( $r < i$ ). The ratio $v_1/v_2$ is constant for any pair of media, giving us the constant refractive index.

Alternative Method — Deriving Reflection Law Similarly

For reflection, both the incident and reflected wavelets travel in the same medium at speed $v$ . Following identical geometry:

\sin i = \frac{v t}{AC}, \quad \sin r' = \frac{v t}{AC}

Therefore $\sin i = \sin r'$ , giving $i = r'$ — the law of reflection.

For CBSE boards, this derivation is a guaranteed 5-mark question. Draw a neat diagram with clearly labelled wavefronts, secondary wavelets, and angles. Examiners give 1-2 marks just for the diagram. Always label $v_1 t$ and $v_2 t$ on your figure.

Common Mistake

Students often mix up which angle is $i$ and which is $r$ in the Huygens construction. Remember: the angle of incidence is between the incident wavefront and the boundary (or equivalently, between the incident ray and the normal). In the wavefront picture, $\sin i = BC/AC$ where BC is the distance the wavefront travels in medium 1 during time $t$ . Drawing a clear, labelled diagram before writing any equation prevents this confusion.

Question

Solution — Step by Step

Why This Works

Alternative Method — Deriving Reflection Law Similarly

Common Mistake

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