Wavefronts and Huygens' principle

A stone in a pond

Drop a stone in still water. A ring of ripples spreads out. All the points on one ring are at the same stage of the wave: they rise and fall together. Now imagine the same for light from a bulb. The surface that joins all the points that are in the same stage of the wave is called a wavefront.

• A small source sends out spherical wavefronts.
• A long line source, such as a slit, sends out cylindrical wavefronts.
• A source very far away gives wavefronts that are almost plane. The light from a star is plane when it reaches us, since we see only a tiny piece of a huge sphere.

A ray is the line that shows the direction in which the wave moves. A ray is at right angles to the wavefront at every point.

Huygens' principle

The Dutch scientist Christiaan Huygens gave a simple rule in 1678.

• Every point on a wavefront is a source of small new waves, called secondary wavelets. They spread forward with the speed of the wave.
• After a time $t$, the new wavefront is the surface that touches all the wavelets. This touching surface is called their envelope.

The wavelets of a plane wavefront give another plane wavefront. The wavelets of a spherical wavefront give a larger sphere. A wavelet seems to spread backwards too, but no backward wave is seen. Fresnel later showed that the wavelets are weak in the backward direction, so the rule gives the forward wave only.

The laws of reflection

A plane wavefront $AB$ falls on a plane mirror. Its rays make the angle of incidence $i$ with the normal. Point $A$ touches the mirror first. Point $B$ reaches the mirror at $C$ after a time $\tau$, so $BC = v\tau$.

In the same time, the wavelet from $A$ spreads over a half circle of radius $AE = v\tau$. The reflected wavefront is the line $CE$ from $C$ that touches this half circle.

1. The two right-angled triangles $ABC$ and $AEC$ share the side $AC$.
2. Also $BC = AE$, since both are $v\tau$.
3. So the triangles are the same shape and size, and the angles at $A$ and $C$ match.

This gives $i = r$. The incident ray, the reflected ray and the normal lie in one plane. These are the two laws of reflection, proved from the wave picture.

The laws of refraction

Now the wave goes from medium 1, where its speed is $v_1$, into medium 2, where its speed is $v_2$. Point $B$ of the wavefront $AB$ needs a time $\tau$ to reach the boundary at $C$, so $BC = v_1\tau$. In that time the wavelet from $A$ travels only $AE = v_2\tau$ in medium 2.

The new wavefront $CE$ is tilted. In the right-angled triangle $ABC$, $\sin i = BC/AC = v_1\tau/AC$. In the triangle $AEC$, $\sin r = AE/AC = v_2\tau/AC$. Dividing,

The speed of light in a medium of refractive index $n$ is $v = c/n$. So $v_1/v_2 = n_2/n_1$, and this gives Snell's law:

Here $c = 3.00 \times 10^{8}$ m/s is the speed of light in vacuum. When the second medium is denser, $v_2 < v_1$ and $r < i$: the wave bends towards the normal.

Set medium 1 to $1.5$ and medium 2 to $1.0$. Raise the angle of incidence. Where does the refracted ray vanish? This is total internal reflection. You met it in the chapter on ray optics.

What changes in a medium, and what does not

The source sets the frequency $f$. A wavefront in medium 1 produces exactly one wavefront in medium 2 for each cycle, so the frequency cannot change. Only the speed changes, to $c/n$. From $v = f\lambda$, the wavelength changes in the same ratio:

Here $\lambda_0$ is the wavelength in vacuum (air is almost the same).

A worked example. Light of wavelength $600$ nm in air enters glass of index $1.50$. Find its speed, wavelength and frequency in the glass.

1. Frequency. In air, $f = \dfrac{c}{\lambda_0} = \dfrac{3.00 \times 10^{8}\ \text{m/s}}{600 \times 10^{-9}\ \text{m}} = 5.00 \times 10^{14}\ \text{Hz}$. It is the same in the glass.
2. Speed. We want $v$, so $v = \dfrac{c}{n} = \dfrac{3.00 \times 10^{8}\ \text{m/s}}{1.50} = 2.00 \times 10^{8}\ \text{m/s}$.
3. Wavelength. $\lambda = \dfrac{\lambda_0}{n} = \dfrac{600\ \text{nm}}{1.50} = 400\ \text{nm}$.
4. Check. $f\lambda = 5.00 \times 10^{14}\ \text{Hz} \times 400 \times 10^{-9}\ \text{m} = 2.00 \times 10^{8}\ \text{m/s}$, which is $v$.

A worked example on the angle. A ray in air meets glass of index $1.50$ at $45^\circ$. Then $\sin r = \dfrac{\sin 45^\circ}{1.50} = 0.471$, so $r = 28.1^\circ$. The ray bends towards the normal.

Other books. A-Level and IB write the same law as $n_1\sin\theta_1 = n_2\sin\theta_2$. Some books give the refractive index as $n_{21} = \sin i/\sin r$. We use $n_1\sin i = n_2\sin r$ everywhere.

Practise with new numbers

Try it in Python

The program takes light of wavelength $600$ nm in air into three materials. It finds the speed, the wavelength and the angle of refraction at $40^\circ$. The frequency is printed once, since it never changes.

Common mistakes

• Measuring the angle from the surface. The angles $i$ and $r$ are measured from the normal, not from the surface. They are also the angles between the wavefronts and the surface.
• Saying the frequency changes in glass. The frequency is fixed by the source. The speed and the wavelength change.
• Thinking that light slows because it is "used up". The speed in a medium is lower because of how the medium responds to the wave. It goes back to $c$ when the light leaves.
• Dividing the angles. Snell's law is for the sines: $\sin i/\sin r$ is constant, not $i/r$.

Remember these

• A wavefront is a surface of equal phase. A ray is at right angles to it.
• Huygens' principle: every point of a wavefront is a source of wavelets, and the new wavefront is their envelope.
• $i = r$ follows from two equal right-angled triangles.
• $n_1\sin i = n_2\sin r$ follows from $v = c/n$.
• In a medium: $f$ is unchanged, $v = c/n$ and $\lambda = \lambda_0/n$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. What is the shape of the wavefront of the light that reaches a telescope from a distant star?

2. Light of frequency $6.0 \times 10^{14}$ Hz enters water of index $1.33$. Find its speed and its wavelength in the water.

3. Huygens' wavelets spread in every direction. Why is no wave seen moving backwards?

4. A ray meets a plane mirror so that it makes $30^\circ$ with the surface of the mirror. What are the angles of incidence and reflection?

References

• OpenStax, University Physics Volume 3, Chapter 1 (The Nature of Light): the Huygens construction and the laws of reflection and refraction.
• NCERT Physics Class 12, Chapter 10 (Wave Optics): Huygens' principle, and the proofs of the laws of reflection and refraction.