Physics · Wave Optics

Refraction and Reflection Using Huygens' Principle

Huygens' principle does more than describe how a wavefront advances; applied at an interface it reproduces the entire ray-optics rulebook. Following NCERT §10.3, this note derives Snell's law and the law of reflection purely from wavefront geometry, shows why a wave slows and bends toward the normal in a denser medium, and locates the critical angle for total internal reflection. The frequency-speed-wavelength distinctions developed here are recurring NEET decision points.

The Wavefront Setup

A wavefront is a surface of constant phase: the locus of points that oscillate together. Energy travels perpendicular to the wavefront, and the direction of that perpendicular is what we call a ray. At a large distance from a source, a small patch of a spherical wavefront is effectively flat, and we treat it as a plane wave. This is the object Huygens' construction works on at an interface.

Huygens' principle states that every point on a wavefront is the source of secondary spherical wavelets that spread with the wave speed of the medium; the new wavefront at a later time is the forward common tangent (envelope) to these wavelets. When the wave crosses from medium 1 into medium 2, the only thing that changes is the radius the wavelets reach in a fixed time, because the speed differs. Holding the time the same across both media while the speeds differ is the entire engine behind the laws of refraction and reflection.

SymbolMeaningDefining relation
v1, v2Wave speed in medium 1 and 2Set by the medium
i, rAngle of incidence, angle of refractionMeasured from the normal
n1, n2Refractive indices$n = c/v$
BCDistance foot of wavefront travels in medium 1 in time $t$$BC = v_1 t$
AERadius of secondary wavelet in medium 2 in time $t$$AE = v_2 t$

Refraction of a Plane Wave

Let the surface $PP'$ separate medium 1 (speed $v_1$) from medium 2 (speed $v_2$). A plane wavefront $AB$ arrives at the interface at angle of incidence $i$. The end $A$ has just reached the interface while the end $B$ still has the distance $BC$ to cover in medium 1. In a time $t$, the foot travels $BC = v_1 t$ to land at $C$. During that same interval, the secondary wavelet launched from $A$ has already entered medium 2 and spread out a radius $AE = v_2 t$.

To find the refracted wavefront we draw a sphere of radius $v_2 t$ centred at $A$ in the second medium and draw the tangent plane from $C$ to that sphere. That tangent, $CE$, is the new refracted wavefront. The refracted ray is the normal to $CE$, making angle $r$ with the interface normal.

Figure 1 · Refraction construction P P' Medium 1 (v₁) Medium 2 (v₂ < v₁) normal A B C BC = v₁t incident ray E AE = v₂t refracted ray i r

A plane wave $AB$ incident at angle $i$ refracts at $PP'$. With $v_2 < v_1$, the wavelet radius $AE = v_2t$ is shorter than $BC = v_1t$, so the refracted wavefront $CE$ tilts and the ray bends toward the normal ($r < i$). Adapted from NCERT Fig. 10.4.

Snell's Law from Geometry

Both triangles $ABC$ and $AEC$ share the hypotenuse $AC$ lying along the interface. In triangle $ABC$ the side opposite angle $i$ is $BC$; in triangle $AEC$ the side opposite angle $r$ is $AE$. Reading the sines directly off the geometry:

$$\sin i = \frac{BC}{AC} = \frac{v_1 t}{AC}, \qquad \sin r = \frac{AE}{AC} = \frac{v_2 t}{AC}$$

Dividing the two and cancelling $t$ and $AC$ gives the central result of the construction:

$$\frac{\sin i}{\sin r} = \frac{v_1}{v_2}$$

If $r < i$ (the ray bends toward the normal), then $v_2 < v_1$: the wave is slower in the second medium. This wave-theory prediction is the exact opposite of the corpuscular model, and it was confirmed experimentally by Foucault in 1850 when light was measured to be slower in water than in air. Introducing the refractive index $n = c/v$, with $n_1 = c/v_1$ and $n_2 = c/v_2$, the relation becomes:

$$n_1 \sin i = n_2 \sin r$$

This is Snell's law, now derived from wavefronts rather than postulated.

H
Build the foundation first

The secondary-wavelet idea used here is set up in full in Huygens' Principle.

Frequency, Speed and Wavelength

Track a single crest. If the crest from $B$ reaches $C$ in time $t$, the crest from $A$ must reach $E$ in the same time $t$. So if $BC$ equals one wavelength $\lambda_1$ in medium 1, then $AE$ equals one wavelength $\lambda_2$ in medium 2. Therefore:

$$\frac{\lambda_1}{\lambda_2} = \frac{BC}{AE} = \frac{v_1}{v_2}$$

The number of wavefronts crossing the interface per second cannot change, so the frequency stays the same. With $v = \nu \lambda$ and $\nu$ fixed, the wavelength scales exactly as the speed. Entering a denser medium ($v_1 > v_2$): speed decreases, wavelength decreases, frequency unchanged.

NEET Trap

Frequency vs. speed vs. wavelength on refraction

Three quantities, three different behaviours, and exam stems mix them deliberately. The colour of light is set by frequency, which is fixed by the source; refraction never changes it. Speed and wavelength both change, and they change in the same ratio $v_1/v_2$.

Lock it in: air → glass means $v\downarrow$, $\lambda\downarrow$, $\nu$ unchanged. Wavefronts bend toward the normal because the wave slows in the denser medium.

QuantityAir → denser mediumReason
Frequency $\nu$UnchangedForced oscillation of atoms takes up source frequency
Speed $v$Decreases$v = c/n$ and $n$ is larger
Wavelength $\lambda$Decreases$\lambda = v/\nu$, $v\downarrow$ with $\nu$ fixed
BendingToward normal$r < i$ since $v_2 < v_1$

Refraction at a Rarer Medium

Reverse the situation: light passes into a rarer medium, so $v_2 > v_1$. The identical construction now produces a wavelet radius $AE = v_2 t$ that is larger than $BC$, so the refracted wavefront tilts the other way and the ray bends away from the normal, giving $r > i$. Snell's law $n_1 \sin i = n_2 \sin r$ still holds.

As $i$ grows, $r$ grows faster and reaches $90^\circ$ at a special angle of incidence called the critical angle $i_c$, defined by:

$$\sin i_c = \frac{n_2}{n_1}$$

When $i = i_c$, $\sin r = 1$ so $r = 90^\circ$ and the refracted wavefront grazes along the interface. For $i > i_c$ no refracted wavefront can be constructed at all, and the wave undergoes total internal reflection.

Figure 2 · Rarer medium and critical angle Medium 1 (denser, v₁) Medium 2 (rarer, v₂ > v₁) i r > i i < i_c i_c r = 90° i = i_c

Into a rarer medium the ray bends away from the normal ($r > i$). At $i = i_c$ the refracted ray grazes the surface; beyond it, total internal reflection. Based on NCERT Fig. 10.5 and §10.3.2.

Reflection of a Plane Wave

For reflection both wavefronts stay in the same medium, so there is a single speed $v$. A plane wave $AB$ strikes a reflecting surface $MN$ at angle $i$. The foot $B$ advances to $C$ in time $t$, covering $BC = vt$. At the same time the wavelet from $A$ spreads a radius $vt$ back into the medium. Drawing the tangent plane $CE$ from $C$ to that sphere gives the reflected wavefront, with $AE = BC = vt$.

Compare triangles $EAC$ and $BAC$: each has a right angle, they share the side $AC$, and $AE = BC$. They are therefore congruent, which forces the angle of incidence $i$ to equal the angle of reflection $r$. That is the law of reflection, obtained without any new assumption beyond the wavelet construction.

Figure 3 · Reflection construction M N A B C BC = vt E AE = vt incident reflected i r

Reflection of plane wave $AB$ at surface $MN$. Since $AE = BC = vt$ and the triangles $EAC$ and $BAC$ are congruent, $i = r$. Adapted from NCERT Fig. 10.6.

One consequence ties the chapter together: along any ray, the total time from a point on the object to the corresponding point on the image is the same. When a convex lens focuses light, the central ray travels a shorter geometric path but moves slower through the thicker glass, so it arrives at the same instant as a ray near the rim. The wavefront picture makes this equal-time principle automatic.

Worked Numbers

Worked Example

Monochromatic light of wavelength 600 nm travels from air ($n_1 \approx 1$) into glass of refractive index $n_2 = 1.5$. Find its speed, wavelength and frequency in glass. Take $c = 3.0 \times 10^8\ \text{m s}^{-1}$.

Speed: $v_2 = c/n_2 = (3.0\times10^8)/1.5 = 2.0\times10^8\ \text{m s}^{-1}$.

Wavelength: $\lambda_2 = \lambda_1 (v_2/v_1) = \lambda_1/n_2 = 600/1.5 = 400\ \text{nm}$.

Frequency: unchanged. $\nu = c/\lambda_1 = (3.0\times10^8)/(600\times10^{-9}) = 5.0\times10^{14}\ \text{Hz}$, the same in glass.

Speed and wavelength fall to two-thirds of their air values; frequency holds steady — exactly the pattern Huygens' construction predicts.

Quick Recap

Refraction and reflection from wavefronts

  • In time $t$ the wavelet radius is $v_2 t$ in medium 2 while the foot covers $v_1 t$ in medium 1; the common tangent is the refracted wavefront.
  • Geometry gives $\dfrac{\sin i}{\sin r} = \dfrac{v_1}{v_2}$, hence $n_1 \sin i = n_2 \sin r$ (Snell's law).
  • Bending toward the normal ($r < i$) means $v_2 < v_1$ — the wave slows in the denser medium.
  • On refraction frequency is unchanged; speed and wavelength change in the same ratio $v_1/v_2 = \lambda_1/\lambda_2$.
  • Rarer medium: ray bends away from normal; critical angle $\sin i_c = n_2/n_1$; beyond it, total internal reflection.
  • For reflection, $AE = BC = vt$ makes triangles congruent, so $i = r$.

NEET PYQ Snapshot — Refraction and Reflection Using Huygens' Principle

No NEET paper has set a question solely on the Huygens construction for refraction or reflection; the items below are concept checks built from NCERT §10.3.

Concept

A plane wave passes from air into glass and the refracted ray bends toward the normal. Which statement is correct about the wave in glass?

  1. Speed increases, frequency unchanged
  2. Speed decreases, wavelength decreases, frequency unchanged
  3. Frequency decreases, wavelength unchanged
  4. Speed decreases, frequency decreases
Answer: (2)

Bending toward the normal means $r < i$, so by $\sin i/\sin r = v_1/v_2$ the speed in glass is smaller. With $\nu$ fixed and $\lambda = v/\nu$, the wavelength also decreases; only frequency stays the same.

Concept

In Huygens' construction for reflection of a plane wave at a plane surface, the law of reflection ($i = r$) follows because:

  1. the incident and reflected wavelets travel different distances
  2. the secondary wavelets have zero amplitude backward
  3. $AE = BC = vt$ makes the two triangles congruent
  4. the frequency changes on reflection
Answer: (3)

Both wavefronts are in the same medium, so the wavelet radius $AE$ equals $BC = vt$. The right-angled triangles $EAC$ and $BAC$ share $AC$ and have equal opposite sides, hence are congruent, forcing $i = r$.

FAQs — Refraction and Reflection Using Huygens' Principle

Common decision points the wavefront derivation settles for NEET.

Why does the frequency of light stay the same on refraction?
Reflection and refraction arise through the interaction of the incident light with the atomic constituents of matter. Atoms behave as oscillators that take up the frequency of the external agency (light) causing forced oscillations, and the frequency of light emitted by a charged oscillator equals its frequency of oscillation. Therefore the frequency of the refracted (and reflected) light equals the frequency of the incident light. In the Huygens picture, the crest from B reaching C and the crest from A reaching E happen in the same time t, so the number of wavefronts crossing the interface per second is conserved.
If a refracted ray bends toward the normal, is the speed of light greater or less in the second medium?
Less. Huygens' construction gives sin i / sin r = v1 / v2. If r < i, that is the ray bends toward the normal, then v2 < v1, so the wave travels slower in the second medium. This is the wave-theory prediction, opposite to the corpuscular model, and it was confirmed by Foucault in 1850 when light was shown to be slower in water than in air.
Does the wavelength of light change when it enters a denser medium?
Yes. From the Huygens construction, lambda1 / lambda2 = v1 / v2. When light enters a denser medium where v decreases, the wavelength decreases in the same proportion while the frequency stays constant. So a ray entering glass slows down and its wavelength shrinks, but its colour (set by frequency) is unchanged.
How is Snell's law obtained from Huygens' principle?
A plane wavefront AB strikes the interface PP'. In time t the foot B travels BC = v1 t in medium 1 while the secondary wavelet from A spreads a radius AE = v2 t in medium 2. From triangles ABC and AEC, sin i = BC/AC and sin r = AE/AC, so sin i / sin r = v1 / v2. Writing n = c/v gives n1 sin i = n2 sin r, which is Snell's law.
How does Huygens' construction prove the law of reflection?
For a plane wave AB incident on a reflecting surface MN, the foot B travels BC = vt while the secondary wavelet from A spreads a radius AE = vt, in the same medium. Triangles EAC and BAC share AC, have a right angle each, and have AE = BC, so they are congruent. Hence the angle of incidence i equals the angle of reflection r, which is the law of reflection.
What is the critical angle in the Huygens treatment of refraction at a rarer medium?
When light goes to a rarer medium (v2 > v1) the refracted wave bends away from the normal, so r > i. The critical angle i_c is defined by sin i_c = n2 / n1. At i = i_c, sin r = 1 so r = 90 degrees; for i > i_c there is no refracted wave and the wave undergoes total internal reflection.