Chapter 10 · Wave Optics

10.1 Huygens' Principle and the Laws of Reflection and Refraction

Treat every point of a wavefront as a tiny source of secondary wavelets, and the new wavefront is simply the surface that touches them all. This one rule — Huygens' principle — explains why light travels in straight lines, reflects with $i=r$ and refracts with $\dfrac{\sin i}{\sin r}=\dfrac{v_1}{v_2}$.

Drop a pebble into a still pond and a ring of ripples spreads out; a little later each ring has grown larger by the same amount everywhere, because every part of it moves at the same speed. Now let the ripples run from deep water into a shallow patch, where water waves travel more slowly. The part of each ring that reaches the shallows first falls behind, the ring kinks, and the whole wave changes direction. That is refraction — and it happens for exactly the same reason to sound entering water, to earthquake waves crossing layers of rock, to surf turning to face a beach, and to light entering glass.

In Chapter 9 we used rays and Snell's law as given rules. In this section we stop treating light as rays and treat it as a wave. Around 1678 the Dutch physicist Christiaan Huygens proposed a simple geometrical rule for how any wave moves forward: every point on a wavefront acts as a source of small secondary wavelets, and a moment later the new wavefront is the surface that touches all of them. With this one idea we will derive the laws of reflection and refraction, show that the refractive index is a ratio of speeds, $n=c/v$, and discover what happens to the frequency and the wavelength of light when it enters glass. Huygens' rule also predicted that light should travel more slowly in water than in air — the opposite of Newton's particle picture — and the experiment that settled the question is one of the turning points of physics.

Wavefronts and Rays

A source of light makes the electric and magnetic fields around it oscillate (Chapter 8). Every point near the source oscillates with the same frequency, but points at different distances lag behind by different amounts. Joining up all the points that are oscillating in step gives a surface.

📘 Definition 10.1 — Wavefront

A wavefront is a surface on which every point of the wave is oscillating in the same phase — for example, the surface on which the wave is at a crest at a given instant. A wavefront moves outward from the source with the speed of the wave. The direction in which the energy travels, at right angles to the wavefront at each point, is a ray.

Three shapes cover almost every case (see Interactive 1):

1. Spherical wavefronts spread from a point source in a uniform medium: the wave reaches every direction equally fast, so points in the same phase lie on a sphere centred on the source. The rays are radial.

2. Cylindrical wavefronts come from a line source such as a long, narrow illuminated slit or a fluorescent tube. Points in the same phase lie on a cylinder round the line.

3. Plane wavefronts are what a spherical wavefront becomes very far from its source. A small part of a huge sphere is almost flat, so light from the Sun or a distant star — or from a point source placed at the focus of a converging lens — arrives as a plane wave with parallel rays.

The energy a point source emits each second is spread over the area $4\pi r^2$ of an ever-larger sphere, so the intensity of a spherical wave falls as $1/r^2$ and, because intensity is proportional to the square of the amplitude, its amplitude falls as $1/r$. For a cylindrical wave the area grows only as $r$, so the intensity falls as $1/r$. A plane wave in a non-absorbing medium keeps its amplitude.

The distance between successive wavefronts that are each at a crest is one wavelength $\lambda$. Two points one behind the other along a ray, a distance $\Delta x$ apart, differ in phase by $$\Delta\phi=\frac{2\pi}{\lambda}\,\Delta x,$$ while all points on one wavefront have phase difference zero. This simple link between path and phase is the key to every interference calculation in Section 10.2.

Huygens' Principle

Knowing the wavefront now, where will it be a short time $\Delta t$ later? Huygens' answer treats each point of the wavefront as if it were itself a little source.

📐 Principle 10.1 — Huygens' principle

1. Every point on a given wavefront acts as a source of secondary wavelets — small spherical waves that spread out in the forward direction with the speed $v$ of the wave in that medium.

2. After a time $\Delta t$ each wavelet has radius $v\,\Delta t$. The new position of the wavefront is the forward envelope of the wavelets: the surface that touches (is tangent to) all of them on the forward side.

The wavelets are assumed to have no backward part, so there is no wave travelling back towards the source. (This assumption was justified by later, fuller theories of waves.)

Apply the rule to the three wavefronts. For a plane wavefront, all the wavelets have the same radius $v\Delta t$ and their centres lie on a plane; the surface touching them is another plane, parallel to the first and a distance $v\Delta t$ ahead. The wave moves straight on — rays remain straight lines in a uniform medium. For a spherical wavefront of radius $R$, the envelope is a sphere of radius $R+v\Delta t$ with the same centre: the wave keeps spreading from the same point. If a plane wave meets a barrier with a gap, only the wavelets from points inside the gap survive; their envelope is flat in the middle but curls round at the edges, so the wave leaks a little into the region that ray optics calls the "geometrical shadow". That leakage is diffraction, the subject of Section 10.3.

✏️ Example 1 — Light entering glass and water
Yellow sodium light has wavelength $589\ \text{nm}$ in vacuum (and, to within $0.03\%$, in air). Using $c=3.0\times10^8\ \text{m/s}$ and the results proved later in this section ($v=c/n$, unchanged frequency), find (a) its frequency, (b) its speed and wavelength in glass of refractive index $1.50$, and (c) its speed and wavelength in water, $n=1.33$.
(a) Frequency $\nu=\dfrac{c}{\lambda}=\dfrac{3.0\times10^8}{589\times10^{-9}}\approx5.09\times10^{14}\ \text{Hz}$. This value is the same in every medium — the source sets it.
(b) Glass $v=\dfrac{c}{n}=\dfrac{3.0\times10^8}{1.50}=2.0\times10^8\ \text{m/s}$; $\lambda'=\dfrac{\lambda}{n}=\dfrac{589}{1.50}\approx393\ \text{nm}$.
(c) Water $v=\dfrac{3.0\times10^8}{1.33}\approx2.26\times10^8\ \text{m/s}$; $\lambda'=\dfrac{589}{1.33}\approx443\ \text{nm}$.
Check $v=\nu\lambda'$: $(5.09\times10^{14})(393\times10^{-9})\approx2.0\times10^8\ \text{m/s}$ ✓. The light in the glass is still yellow: the colour we see is fixed by the frequency, not by the wavelength inside the glass.
Interactive 1 — Huygens' Construction, Step by Step

The thick line is the present wavefront. Each dot on it sends out a secondary wavelet whose radius grows as $v\,\Delta t$; when the wavelets have grown fully, the curve touching all of them (drawn in red) becomes the new wavefront, and the construction repeats. Choose the kind of wavefront, and watch what happens at the edges of a gap in a barrier.

step—
wavelet radius $v\,\Delta t$—
new wavefront—

Refraction of a Plane Wave

Let a plane wavefront travel in medium 1, where the wave speed is $v_1$, towards a flat boundary with medium 2, where the speed is $v_2$. The rays make an angle $i$ with the normal to the boundary, so the wavefront — perpendicular to the rays — makes the same angle $i$ with the boundary itself. One edge of the wavefront, $A$, touches the boundary first; the other edge, $B$, still has a distance $BC$ to go before it reaches the boundary at $C$.

ABCEir BC = v₁τAE = v₂τmedium 1, speed v₁medium 2, speed v₂ < v₁incident wavefront AB (red)refracted wavefront CE (blue)
Huygens' construction for refraction. While edge $B$ travels $BC=v_1\tau$ to the boundary, the wavelet from $A$ grows to radius $AE=v_2\tau$; wavelets from points between $A$ and $C$ (one shown) start later and are smaller. The refracted wavefront $CE$ touches them all.

Let the time taken by $B$ to reach $C$ be $\tau$, so $BC=v_1\tau$. During the same time the secondary wavelet sent into medium 2 from $A$ has grown to radius $v_2\tau$. A point of the boundary halfway between $A$ and $C$ is reached halfway through the interval, so its wavelet has had only half as long to grow, and so on: the wavelets shrink steadily to zero radius at $C$. The forward envelope is therefore the straight line from $C$ that just touches the wavelet from $A$, at the point $E$, with $AE=v_2\tau$. That line, $CE$, is the refracted wavefront, and the refracted ray $AE$ is perpendicular to it.

Triangles $ABC$ and $AEC$ are both right-angled (at $B$ and at $E$) and share the hypotenuse $AC$. The angle of the incident wavefront to the boundary is $\angle BAC=i$; the angle of the refracted wavefront to the boundary is $\angle ACE=r$, which is also the angle the refracted ray makes with the normal. So $$\sin i=\frac{BC}{AC}=\frac{v_1\tau}{AC},\qquad \sin r=\frac{AE}{AC}=\frac{v_2\tau}{AC},$$ and dividing, $$\frac{\sin i}{\sin r}=\frac{v_1}{v_2}.$$ The ratio is a constant for the two media, whatever the angle — that is Snell's law, now derived rather than assumed. If $v_2\lt v_1$, then $r\lt i$: the ray bends towards the normal on entering the slower medium.

📐 Result 10.2 — Refraction from Huygens' principle (Snell's law)

A plane wave crossing a plane boundary from medium 1 (wave speed $v_1$) into medium 2 (wave speed $v_2$) obeys

$$\frac{\sin i}{\sin r}=\frac{v_1}{v_2}=n_{21},$$

where $i$ and $r$ are the angles the rays make with the normal, the incident ray, the refracted ray and the normal lie in one plane, and $n_{21}$ is the refractive index of medium 2 relative to medium 1. Defining the (absolute) refractive index of a medium as $n=\dfrac{c}{v}$, with $c$ the speed of light in vacuum,

$$n_{21}=\frac{n_2}{n_1}\qquad\text{and}\qquad n_1\sin i=n_2\sin r.$$

A medium of larger $n$ is optically denser: light travels more slowly in it.

What happens to frequency and wavelength? In time $\tau$ the incident wavefront moves forward by $v_1\tau$ and the refracted one by $v_2\tau$, and every point of the boundary is set oscillating by the incident wave and passes the same oscillation on to medium 2. The boundary cannot pile up or lose crests, so the number of crests arriving per second equals the number leaving per second: the frequency does not change. Since $v=\nu\lambda$ in each medium, the wavelength changes in proportion to the speed.

📐 Result 10.3 — Frequency, speed and wavelength in a medium

When light passes from one medium into another, its frequency $\nu$ stays the same; its speed and wavelength change in the same ratio:

$$\frac{\lambda_1}{\lambda_2}=\frac{v_1}{v_2}=\frac{n_2}{n_1}.$$

In particular, light of vacuum wavelength $\lambda$ has, in a medium of refractive index $n$, speed $v=\dfrac{c}{n}$ and wavelength $\lambda'=\dfrac{\lambda}{n}$.

Going from a denser to a rarer medium. Now $v_2\gt v_1$: the wavelet from $A$ grows faster than $B$ moves, so $AE\gt BC$ and $r\gt i$ — the ray bends away from the normal. As $i$ grows, $AE=AC\sin i\cdot\dfrac{v_2}{v_1}$ eventually equals $AC$, giving $r=90^\circ$; this happens at the critical angle, $\sin i_c=\dfrac{v_1}{v_2}=\dfrac{n_2}{n_1}$. For larger $i$ the wavelet from $A$ is bigger than $AC$ itself, no line from $C$ can touch it, and there is no refracted wave at all: the light is totally internally reflected, as in Chapter 9 — now explained by the wave picture.

✏️ Example 2 — Carrying out the construction
A plane wavefront $6.0$ cm wide (the distance $AB$ measured along the wavefront) travels in air and meets a flat water surface at an angle of incidence of $40^\circ$. Take $n_\text{water}=1.33$ and $c=3.0\times10^8\ \text{m/s}$. Find (a) the length $AC$ of the surface the beam covers, (b) the distance $BC$ the far edge must still travel when edge $A$ touches the water, and the time this takes, (c) the radius $AE$ of the wavelet from $A$ when $B$ arrives at $C$, (d) the angle of refraction, and (e) the width of the refracted wavefront $CE$.
(a) AC In the right triangle $ABC$, $AB=AC\cos i$: $AC=\dfrac{6.0}{\cos40^\circ}=\dfrac{6.0}{0.766}\approx7.83$ cm.
(b) BC and τ $BC=AC\sin i=7.83\times0.643\approx5.03$ cm. Time $\tau=\dfrac{BC}{c}=\dfrac{0.0503}{3.0\times10^8}\approx1.7\times10^{-10}$ s.
(c) AE In water $v_2=c/1.33$, so $AE=v_2\tau=\dfrac{BC}{1.33}=\dfrac{5.03}{1.33}\approx3.79$ cm.
(d) r $\sin r=\dfrac{AE}{AC}=\dfrac{3.79}{7.83}\approx0.483$, so $r\approx28.9^\circ$.
(e) CE $CE=AC\cos r=7.83\times0.875\approx6.86$ cm — the refracted beam is wider than the incident one, because it leaves the surface closer to the normal.
Check Snell's law directly: $\sin r=\dfrac{\sin40^\circ}{1.33}=\dfrac{0.643}{1.33}=0.483$ ✓. Set Interactive 2 to water and $i=40^\circ$: it reports $r=28.9^\circ$ from the construction.
✏️ Example 3 — From glass into water
A ray in glass ($n=1.50$) meets a glass–water boundary ($n=1.33$) at $30^\circ$ to the normal. (a) Find the angle of refraction in the water. (b) How many times faster does light travel in the water than in the glass? (c) Above what angle of incidence in the glass is there no refracted wave?
(a) Snell $1.50\sin30^\circ=1.33\sin r$: $\sin r=\dfrac{0.750}{1.33}\approx0.564$, $r\approx34.3^\circ$ — bent away from the normal, into the faster medium.
(b) Speeds $\dfrac{v_w}{v_g}=\dfrac{n_g}{n_w}=\dfrac{1.50}{1.33}\approx1.13$.
(c) Critical angle $\sin i_c=\dfrac{1.33}{1.50}\approx0.887$, so $i_c\approx62.5^\circ$. Beyond it the wavelet from $A$ in the water is longer than $AC$ and no refracted wavefront can be drawn.
Interactive 2 — A Wavefront Crossing a Boundary

A train of plane wavefronts (crests, one wavelength apart) comes down from the top. Each point of the boundary starts its own wavelet in the lower medium the moment the leading crest reaches it (purple circles); the refracted crest is the line touching them all. Because the wave is slower below, the crests crowd closer together and the train turns towards the normal. Change the angle of incidence and the media; choose a denser upper medium to see total internal reflection. Tick the box to draw the reflected wavefront as well.

40°
$r$ from the construction—
$r$ from $n_1\sin i=n_2\sin r$—
$\lambda_2/\lambda_1=v_2/v_1$—
$\lambda$ in each medium (600 nm in vacuum)—

Reflection of a Plane Wave, and Wavefronts through Prisms, Lenses and Mirrors

Huygens' construction works just as well for reflection. Let a plane wavefront $AB$ fall on a plane mirror at angle of incidence $i$, with edge $A$ touching the mirror first. While edge $B$ travels the distance $BC=v\tau$ to the mirror, the wavelet from $A$ spreads back into the same medium, so it has the same speed and reaches radius $AE=v\tau$. The reflected wavefront is the line $CE$ from $C$ touching that wavelet (and all the smaller ones from points between $A$ and $C$).

The right triangles $ABC$ (right angle at $B$) and $CEA$ (right angle at $E$) share the hypotenuse $AC$ and have $BC=AE$. They are congruent, so $\angle BAC=\angle ECA$: the incident and the reflected wavefronts make equal angles with the mirror. Since the rays are perpendicular to the wavefronts, the rays make equal angles with the normal.

📐 Law 10.4 — Reflection from Huygens' principle

When a plane wave is reflected at a plane surface, the angle of reflection equals the angle of incidence, $r=i$, and the incident ray, the reflected ray and the normal lie in one plane. The reflected wave has the same speed, frequency and wavelength as the incident wave.

Prisms, lenses and mirrors in the wave picture. Huygens' principle gives every result of ray optics a second, very physical meaning: a lens or a mirror works by reshaping wavefronts.

1. A thin prism. A plane wavefront falls on a prism. The part of it that crosses the thick base spends longer in the slow glass than the part that crosses near the thin apex, so it falls behind. The emerging wavefront is still plane, but tilted towards the base: the light is deviated towards the base.

2. A convex lens. The centre of a convex lens is thicker than its edges, so the middle of a plane wavefront is delayed the most. The emerging wavefront is a sphere curving round a point beyond the lens — it converges to the focus.

3. A concave mirror. The edges of a plane wavefront reach the mirror first and are reflected first, so after reflection the wavefront is again a converging sphere, centred on the focus.

In each case every point of the emerging wavefront is in the same phase, so the light takes the same time from one wavefront to the next along every ray. For a lens this means the time from a point object to its image is the same along every ray: the central ray travels a shorter distance but more of it in slow glass; the edge ray travels further but mostly in fast air. That equality of time is why all the rays from the object arrive at the image in step.

✏️ Example 4 — A wavefront held back by a glass plate
A plane wavefront of light of vacuum wavelength $500\ \text{nm}$ meets a glass plate $2.0$ mm thick ($n=1.50$) at normal incidence. Part of the wavefront passes through the plate; the rest travels the same $2.0$ mm in air. (a) How long does the light take to cross the plate? (b) By how much is that part of the wavefront delayed compared with the part in air? (c) How many wavelengths fit into the plate, and into the same thickness of air?
(a) In glass $v=\dfrac{c}{n}=2.0\times10^8\ \text{m/s}$, so $t_g=\dfrac{2.0\times10^{-3}}{2.0\times10^8}=1.0\times10^{-11}$ s.
(b) Delay In air $t_a=\dfrac{2.0\times10^{-3}}{3.0\times10^8}\approx6.7\times10^{-12}$ s, so the glass part lags by $\Delta t\approx3.3\times10^{-12}$ s. The emerging wavefront has a "step" in it, $c\,\Delta t=(n-1)\times2.0\ \text{mm}=1.0$ mm behind.
(c) Wavelengths In glass $\lambda'=\dfrac{500}{1.50}\approx333\ \text{nm}$, so the plate holds $\dfrac{2.0\times10^{-3}}{333\times10^{-9}}=6000$ wavelengths; the same thickness of air holds $\dfrac{2.0\times10^{-3}}{500\times10^{-9}}=4000$.
Lesson The glass part of the wavefront is $2000$ whole wavelengths behind. A tiny change in thickness or in $n$ would shift that by a fraction of a wavelength — Section 10.2 shows how interference detects such shifts.
✏️ Example 5 — Particles or waves? The speed of light in water
Newton explained refraction by supposing that light particles are pulled towards a denser medium as they cross its surface, which increases the component of their velocity perpendicular to the surface while leaving the parallel component unchanged. (a) Show that this model gives $\dfrac{\sin i}{\sin r}=\dfrac{v_2}{v_1}$. (b) For water ($n=1.33$), what speed of light in water does each model predict? (c) How was the question settled?
(a) Particle model The parallel component is unchanged: $v_1\sin i=v_2\sin r$, so $\dfrac{\sin i}{\sin r}=\dfrac{v_2}{v_1}$ — the speed ratio is upside down compared with Result 10.2.
(b) Predictions Both models must give the measured $\dfrac{\sin i}{\sin r}=1.33$ for air to water. Particles: $v_\text{water}=1.33c\approx4.0\times10^8\ \text{m/s}$ (faster). Waves: $v_\text{water}=\dfrac{c}{1.33}\approx2.26\times10^8\ \text{m/s}$ (slower).
(c) Experiment In 1850 Léon Foucault, using a rapidly rotating mirror, compared the speed of light in air and in water directly and found it slower in water — about $3/4$ of its speed in air, as the wave model predicts.
✏️ Example 6 — Sound entering water
Huygens' principle applies to every kind of wave. Sound travels at about $340$ m/s in air and $1480$ m/s in water. (a) A sound wave in air meets a calm lake at $10^\circ$ to the normal. At what angle does it travel in the water? (b) Beyond what angle of incidence in air can no sound enter the water? (c) Is the water optically "denser" or "rarer" for sound?
(a) Refraction $\sin r=\dfrac{v_2}{v_1}\sin i$ $=\dfrac{1480}{340}\times\sin10^\circ$ $=4.35\times0.174\approx0.756$, so $r\approx49^\circ$ — bent far away from the normal.
(b) Critical angle $r=90^\circ$ when $\sin i=\dfrac{340}{1480}\approx0.230$, i.e. $i\approx13.3^\circ$. Sound arriving more obliquely than this is totally reflected at the surface.
(c) Interpretation For sound, water is the faster medium, so it behaves like a "rarer" medium — the opposite of light. "Denser" in optics means "slower", not "heavier".
Check The construction needs $AE=AC\sin r\le AC$: at $i=13.3^\circ$, $AE=4.35\times AC\sin13.3^\circ\approx AC$ ✓.
⚠️ Common mistakes with wavefronts and Huygens' principle

1. Confusing rays and wavefronts. Rays are perpendicular to wavefronts. The angle of incidence is the angle between the ray and the normal — which equals the angle between the wavefront and the surface.

2. Thinking the frequency changes on refraction. Speed and wavelength change; frequency (and so colour) does not.

3. Writing $\dfrac{\sin i}{\sin r}=\dfrac{v_2}{v_1}$. That is the particle model's prediction, which experiment rejects. The correct ratio is $\dfrac{v_1}{v_2}$: the ray bends towards the normal in the slower medium.

4. Drawing the wavelets in the wrong medium. For refraction the wavelets from the boundary grow in medium 2 with speed $v_2$; for reflection they grow back in medium 1 with speed $v_1$.

5. Forgetting that the wavelets start at different times. The wavelet from $A$ has had the whole time $\tau$ to grow; the one from $C$ has only just started. That is why the envelope is a tilted straight line.

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — optical path length

Not required by the CBSE syllabus. Huygens' principle and interference are listed by JEE Main, NEET and JEE Advanced; none names optical path length, but interference problems with sheets and slabs rely on it.

A distance $L$ in a medium of refractive index $n$ takes light the time $\dfrac{L}{c/n}=\dfrac{nL}{c}$ — the same time it would take to travel the distance $nL$ in vacuum. The product $nL$ is called the optical path length. Phase differences depend on optical paths: two points a distance $L$ apart along a ray in the medium differ in phase by $\dfrac{2\pi}{\lambda'}L=\dfrac{2\pi}{\lambda}(nL)$, where $\lambda$ is the vacuum wavelength.

A sheet of thickness $t$ and index $n$ placed across part of a beam replaces a path $t$ of air by an optical path $nt$, so it adds an extra optical path $(n-1)t$ — exactly the $1.0$ mm step of Example 4. The equal-time property of lenses becomes: all rays from an object point to its image have equal optical paths. Fermat's principle (light follows the path of stationary — usually least — time between two points) is the ray-optics form of the same idea, and it also yields $n_1\sin i=n_2\sin r$. JEE problems use the extra optical path $(n-1)t$ to find how far a thin film shifts interference fringes (Section 10.2).

Exercises

I
Wavefronts and Huygens' principle.
  • 1 State the shape of the wavefront (a) at a small distance from a point source, (b) emerging from a long, narrow illuminated slit, (c) of starlight arriving at the Earth.
  • 2 What is the angle between a ray and the wavefront at the point where they meet?
  • 3 For the spherical wave from a point source in a non-absorbing medium, find the ratio of the intensity, and of the amplitude, at $2.0$ m from the source to their values at $1.0$ m.
  • 4 A long fluorescent tube acts as a line source. Compare the intensities, and the amplitudes, of its light at $2.0$ m and at $8.0$ m from the tube (ignore absorption and end effects).
  • 5 Huygens' construction would also produce a wavefront travelling backwards, towards the source. How does the principle deal with this?
  • 6 What is the phase difference between (a) two points on the same wavefront, (b) two points $150$ nm apart along a ray of light of wavelength $600$ nm?
  • 7 Sunlight reaches the Earth as part of a spherical wavefront of radius $1.5\times10^{11}$ m. Across a window $1.0$ m wide, by how much does this wavefront depart from a flat plane? (Use the sag $s\approx\dfrac{x^2}{2R}$ for a half-width $x$.)
1 (a) Spherical; (b) cylindrical; (c) plane (a tiny part of an enormous sphere).
2 $90^\circ$ — a ray is the normal to the wavefront.
3 $I\propto1/r^2$: intensity ratio $\tfrac14$. Amplitude $\propto1/r$: ratio $\tfrac12$.
4 For a cylindrical wave $I\propto1/r$: $\dfrac{I_2}{I_8}=\dfrac{8.0}{2.0}=4$. Amplitude $\propto1/\sqrt r$: ratio $2$.
5 It assumes the secondary wavelets act only in the forward direction (zero amplitude backwards), so only the forward envelope is taken. Fuller wave theories later justified this.
6 (a) Zero, by definition of a wavefront. (b) $\Delta\phi=\dfrac{2\pi}{\lambda}\Delta x=\dfrac{2\pi\times150}{600}=\dfrac{\pi}{2}$ ($90^\circ$).
7 $x=0.50$ m: $s\approx\dfrac{(0.50)^2}{2(1.5\times10^{11})}\approx8.3\times10^{-13}$ m — a millionth of a wavelength. For every practical purpose sunlight arrives as a plane wave.
II
Speed, frequency and wavelength in a medium. $c=3.0\times10^8$ m/s.
  • 1 Find the speed of light in diamond, $n=2.42$.
  • 2 A helium–neon laser emits light of wavelength $633$ nm in air. Find its frequency and its wavelength in water ($n=1.33$).
  • 3 Light has wavelength $400$ nm inside glass of refractive index $1.50$. What is its wavelength in air? Does it look violet inside the glass?
  • 4 Light travels at $2.25\times10^8$ m/s in a liquid. Find the refractive index of the liquid.
  • 5 Find the refractive index of glass ($n=1.50$) relative to water ($n=1.33$).
  • 6 How long does light take to cross a fish tank $10$ cm wide filled with water ($n=1.33$)? Ignore the glass walls.
  • 7 How many wavelengths of light of vacuum wavelength $500$ nm fit into a glass sheet $1.0$ mm thick ($n=1.50$)? How many into $1.0$ mm of air?
  • 8 Light passes from air into glass of refractive index $1.6$. Which of its speed, frequency and wavelength change, and by what factor?
1 $v=\dfrac{3.0\times10^8}{2.42}\approx1.24\times10^8$ m/s.
2 $\nu=\dfrac{3.0\times10^8}{633\times10^{-9}}\approx4.74\times10^{14}$ Hz; $\lambda'=\dfrac{633}{1.33}\approx476$ nm.
3 $\lambda=n\lambda'=1.50\times400=600$ nm (orange). It does not look violet: the frequency, which fixes the colour, is that of $600$ nm light in air.
4 $n=\dfrac{c}{v}=\dfrac{3.0\times10^8}{2.25\times10^8}\approx1.33$ (water, for instance).
5 $n_{gw}=\dfrac{1.50}{1.33}\approx1.13$.
6 $t=\dfrac{nL}{c}=\dfrac{1.33\times0.10}{3.0\times10^8}\approx4.4\times10^{-10}$ s.
7 In glass $\lambda'=333$ nm: $\dfrac{1.0\times10^{-3}}{333\times10^{-9}}=3000$. In air: $\dfrac{1.0\times10^{-3}}{500\times10^{-9}}=2000$.
8 Frequency unchanged. Speed and wavelength are both divided by $1.6$ (each becomes $0.625$ of its value in air).
III
Huygens' constructions for refraction and reflection.
  • 1 A plane wave in air meets glass ($n=1.50$) at $i=60^\circ$. Find the angle of refraction.
  • 2 In a Huygens construction for air to a medium of $n=1.6$, the far edge of the incident wavefront travels $4.0$ cm to reach the boundary. What is the radius of the wavelet from the near edge at that moment?
  • 3 In a construction, the beam covers $AC=8.0$ cm of the boundary; the edge $B$ travels $BC=5.0$ cm while the wavelet from $A$ grows to $AE=3.0$ cm. Find $i$, $r$ and the relative refractive index $n_{21}$.
  • 4 A plane wavefront makes an angle of $25^\circ$ with a plane mirror as it arrives. What is the angle of incidence, and what angle does the reflected wavefront make with the mirror?
  • 5 Light in water ($n=1.33$) meets the water–air surface at $40^\circ$ to the normal. Find the angle of refraction in air.
  • 6 Light in glass ($n=1.50$) meets a glass–air surface at $45^\circ$. Use the Huygens construction to explain what happens.
  • 7 Describe the emerging wavefront when a plane wavefront passes through (a) a thin prism, (b) a thin convex lens, (c) a thin concave lens, and when it is reflected by (d) a concave mirror.
1 $\sin r=\dfrac{\sin60^\circ}{1.50}=\dfrac{0.866}{1.50}\approx0.577$, $r\approx35.3^\circ$.
2 $AE=v_2\tau=\dfrac{v_2}{v_1}BC=\dfrac{4.0}{1.6}=2.5$ cm.
3 $\sin i=\dfrac{5.0}{8.0}=0.625$, $i\approx38.7^\circ$; $\sin r=\dfrac{3.0}{8.0}=0.375$, $r\approx22.0^\circ$; $n_{21}=\dfrac{v_1}{v_2}=\dfrac{BC}{AE}=\dfrac{5.0}{3.0}\approx1.67$.
4 The angle between wavefront and surface equals the angle between ray and normal: $i=25^\circ$. The reflected wavefront also makes $25^\circ$ with the mirror (on the other side of the normal).
5 $\sin r=1.33\sin40^\circ\approx0.855$, $r\approx58.7^\circ$ — away from the normal.
6 The wavelet from $A$ in air would need radius $AE=\dfrac{v_\text{air}}{v_\text{glass}}BC=1.50\,AC\sin45^\circ\approx1.06\,AC$, longer than $AC$, so no line from $C$ can touch it: there is no refracted wavefront and the light is totally internally reflected (the critical angle is $\sin^{-1}(1/1.50)\approx41.8^\circ$).
7 (a) Plane, but tilted towards the base (the base part is delayed most). (b) Spherical, converging to the focus (centre delayed most). (c) Spherical, diverging from the virtual focus (edges delayed most). (d) Spherical, converging to the focus.
IV
Applications and longer problems.
  • 1 Sound travels at $340$ m/s in air and $1500$ m/s in sea water. Find the largest angle of incidence at which sound from the air can still enter the sea.
  • 2 Modelling. Earthquake P-waves travel at $6.0$ km/s in the Earth's crust and $8.0$ km/s in the mantle below. A wave in the crust meets the boundary at $30^\circ$ to the normal. Find its direction in the mantle and the critical angle for the crust–mantle boundary.
  • 3 Ocean waves travel at $10$ m/s in deep water and $5.0$ m/s in shallower water near a beach. Wave crests approach the shallows with their rays at $60^\circ$ to the normal. At what angle do they travel in the shallow water? Explain why waves tend to arrive parallel to the shore.
  • 4 How long does a light pulse take to travel along a $1.0$ km optical fibre whose core has $n=1.48$? (Assume it travels along the axis.)
  • 5 By how much is a light pulse delayed when a glass plate $5.0$ mm thick ($n=1.50$) is placed in its path at normal incidence?
  • 6 On a hot day, the air just above a road is hotter, and so has a slightly smaller refractive index, than the air higher up. Use Huygens' principle to explain why light from the sky travelling nearly horizontally near the road curves upwards, producing a "wet road" mirage.
1 $\sin i_c=\dfrac{340}{1500}\approx0.227$, $i_c\approx13.1^\circ$.
2 $\sin r=\dfrac{8.0}{6.0}\sin30^\circ\approx0.667$, $r\approx41.8^\circ$ (away from the normal). Critical angle: $\sin i_c=\dfrac{6.0}{8.0}=0.75$, $i_c\approx48.6^\circ$.
3 $\sin r=\dfrac{5.0}{10}\sin60^\circ\approx0.433$, $r\approx25.7^\circ$. As the water gets shallower the waves keep slowing, and the rays keep turning towards the normal, so the crests become nearly parallel to the shoreline.
4 $t=\dfrac{nL}{c}=\dfrac{1.48\times1000}{3.0\times10^8}\approx4.9\times10^{-6}$ s $=4.9\ \mu\text{s}$.
5 $\Delta t=\dfrac{(n-1)L}{c}=\dfrac{0.50\times5.0\times10^{-3}}{3.0\times10^8}\approx8.3\times10^{-12}$ s.
6 The lower part of a (nearly vertical) wavefront is in hotter air, where light is faster, so in each instant its wavelets are larger than those higher up. The wavefront swings round, its lower end moving ahead, and the ray — perpendicular to it — curves upwards. Light from the sky reaches the eye from below the horizon, and the brain sees an image of the sky on the road, like a reflection from water.
🎯 Exam corner — Huygens' principle, reflection and refraction BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks State Huygens' principle. Using it, draw the refracted wavefront for a plane wave incident obliquely on the plane boundary between a rarer and a denser medium, and hence verify Snell's law.

Principle Every point of a wavefront is a source of secondary wavelets spreading forward with the wave speed; the new wavefront is the forward envelope (common tangent) of the wavelets.
Diagram Incident wavefront $AB$ at angle $i$ to the boundary; edge $B$ travels $BC=v_1\tau$; the wavelet from $A$ in medium 2 has radius $AE=v_2\tau$; the refracted wavefront is the tangent $CE$ from $C$, at angle $r$ to the boundary.
Proof Right triangles $ABC$ and $AEC$ share hypotenuse $AC$: $\sin i=\dfrac{v_1\tau}{AC}$, $\sin r=\dfrac{v_2\tau}{AC}$, so $\dfrac{\sin i}{\sin r}=\dfrac{v_1}{v_2}=\dfrac{n_2}{n_1}$, a constant — Snell's law. Since $v_2\lt v_1$, $r\lt i$.
Exam tip Label $BC=v_1\tau$ and $AE=v_2\tau$ on the diagram and mark the right angles at $B$ and $E$ — most of the marks are for the diagram.

E2. Board · 1 mark Assertion (A): When light passes from air into glass, its frequency remains unchanged. Reason (R): The frequency of a wave is fixed by its source, and every point of the boundary oscillates at the frequency of the incident wave. (A) Both A and R are true and R is the correct explanation of A. (B) Both are true but R is not the correct explanation of A. (C) A is true, R is false. (D) A is false, R is true.

Concept The boundary is driven by the incident wave and passes the same oscillation on; crests cannot pile up or vanish at the boundary. Speed and wavelength change, frequency does not.
Answer (A). Exam tip: colour is set by frequency, which is why a glass prism or a swimming pool does not change colours.

E3. JEE Main Light of wavelength $600$ nm in air enters a medium of refractive index $1.5$. Its wavelength and frequency in the medium are (A) $400$ nm, $5.0\times10^{14}$ Hz (B) $900$ nm, $5.0\times10^{14}$ Hz (C) $400$ nm, $3.3\times10^{14}$ Hz (D) $600$ nm, $5.0\times10^{14}$ Hz.

Equations $\nu=\dfrac{c}{\lambda}=\dfrac{3.0\times10^8}{600\times10^{-9}}=5.0\times10^{14}$ Hz, unchanged; $\lambda'=\dfrac{\lambda}{n}=400$ nm.
Answer (A). Exam tip: (C) divides the frequency by $n$ as well — only one of $\nu$ and $\lambda$ can change, and it is $\lambda$.

E4. NEET The speed of light is $2.0\times10^8$ m/s in medium A and $2.5\times10^8$ m/s in medium B. The critical angle for light going from A to B is (A) $\sin^{-1}\tfrac45$ (B) $\sin^{-1}\tfrac35$ (C) $\tan^{-1}\tfrac45$ (D) there is no critical angle, because B is denser.

Concept Light is slower in A, so A is the denser medium and total internal reflection is possible going from A to B. At the critical angle $r=90^\circ$: $\sin i_c=\dfrac{v_A}{v_B}=\dfrac{2.0}{2.5}=\dfrac45$.
Answer (A). Exam tip: "denser" means slower: check the speeds, not the names of the media.

E5. JEE Advanced (One or more correct.) A parallel beam of light passes obliquely from medium 1 into medium 2, in which the wave speed is smaller. Then (A) the frequency is the same in both media; (B) the wavelength in medium 2 is $\lambda_1v_2/v_1$; (C) the refracted beam is wider (measured across the beam) than the incident beam; (D) the refracted ray bends away from the normal.

Options (A) true (Result 10.3). (B) true: $\lambda\propto v$ at fixed $\nu$. (C) The beam covers a length $AC$ of the boundary; its width is $AC\cos i$ before and $AC\cos r$ after. Since $r\lt i$, $\cos r\gt\cos i$: true. (D) false — it bends towards the normal.
Answer (A), (B), (C). Exam tip: (C) is the classic trap; Example 2 shows a $6.0$ cm beam becoming $6.86$ cm wide in water.

E6. Board · Case study In a school ripple tank, a vibrating bar makes straight water waves of frequency $10$ Hz. They travel at $0.30$ m/s over the deep part of the tank and at $0.20$ m/s over a glass plate that makes the water shallow. (i) (2 marks) Find the wavelength over each part. (ii) (1 mark) The wave rays meet the edge of the plate at $45^\circ$ to the normal. Find the angle of refraction. (iii) (1 mark) Can waves travelling from the shallow part into the deep part be totally reflected? If so, find the critical angle.

(i) $\lambda=\dfrac{v}{\nu}$: deep $\dfrac{0.30}{10}=3.0$ cm; shallow $\dfrac{0.20}{10}=2.0$ cm (same frequency).
(ii) $\sin r=\dfrac{0.20}{0.30}\sin45^\circ\approx0.471$, $r\approx28.1^\circ$.
(iii) Yes — the shallow part is the slower ("denser") medium. $\sin i_c=\dfrac{0.20}{0.30}$, $i_c\approx41.8^\circ$. Exam tip: Huygens' principle is not just for light; case studies often use water or sound waves.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Why do ocean waves almost always arrive parallel to the beach, even when the wind drives them in at an angle — and where along a coast does wave energy concentrate?

Why it matters. Coastal engineers in the Netherlands, Japan, Australia and India design harbours, breakwaters and sea walls around wave refraction; surf forecasters use it to predict where waves break best; and tsunami warning centres use the same physics to predict where a tsunami will strike hardest. In water much shallower than the wavelength, the wave speed depends only on the depth $h$: $v\approx\sqrt{gh}$. As waves move into shallower water they slow down and, by Huygens' principle, turn.

What to do.

  1. Using $v=\sqrt{gh}$ with $g=9.8\ \text{m/s}^2$, find the wave speed at depths of $20$, $10$, $5$ and $2$ m.
  2. A wave ray makes $50^\circ$ with the normal to the depth contours where the water is $20$ m deep. Treat the sea floor as a series of steps parallel to the shore and apply $\dfrac{\sin\theta}{v}=\text{constant}$ at each step. Find the direction of the ray at $10$, $5$ and $2$ m depth.
  3. Sketch the wave crests (perpendicular to the rays) as they approach the shore.
  4. Now imagine a headland: the sea floor rises towards a point jutting out into the sea. Sketch how the crests bend round it, and explain why wave energy concentrates on headlands and spreads out in bays.

Tools: Calculator or spreadsheet; graph paper. A coastal chart with depth contours (any freely available nautical chart) makes step 4 concrete.

Hand in: The table of speeds and angles, the crest sketch and a paragraph on headlands and bays.

Going further. A tsunami has a wavelength of a hundred kilometres or more, so even the deep ocean ($4000$ m) counts as "shallow" for it. Estimate its speed in the open ocean and compare it with an airliner.

Instructor note: $v\approx14.0,\ 9.9,\ 7.0,\ 4.4$ m/s at $20,\ 10,\ 5,\ 2$ m. $\sin\theta/v=\sin50^\circ/14.0\approx0.0547$ s/m, giving $\theta\approx32.8^\circ,\ 22.5^\circ,\ 14.0^\circ$ at $10,\ 5,\ 2$ m: the crests swing round to within $14^\circ$ of the shoreline. Headlands: crests in the shallower water off the point slow first, so the rays converge on the headland (energy focusing); in bays they diverge. Tsunami: $\sqrt{9.8\times4000}\approx198$ m/s $\approx710$ km/h, comparable with a jet airliner.

🧪 Self-check quiz

0/5 correct

Q1. Light from a very distant star reaches a telescope. The wavefronts entering the telescope are

Q2. When a light wave passes from air into glass, the quantity that does not change is its

Q3. In Huygens' construction, the new wavefront is

Q4. The wave theory and the particle theory of light both explain refraction into water. The wave theory predicts that light in water travels

Q5. Light of wavelength $600$ nm in air has wavelength $400$ nm in a liquid. The refractive index of the liquid is