Wave Optics

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Physics Class XII, Chapter 10.

Is light a stream of particles or a wave?

Isaac Newton favoured particles. Christiaan Huygens argued for waves. For more than a century Newton's reputation tipped the argument, until a test came along. If light is a stream of corpuscles, then entering water pulls them in and speeds them up. If light is a wave, the wave slows down in water. In 1850 Léon Foucault measured the speed of light in water and found it smaller than in air. The wave model had won.

In 1801 Thomas Young produced interference of light, something only waves can do. Later Maxwell showed that light is an electromagnetic wave. Everything in this chapter rests on that.

Wavefronts and Huygens' principle

A wavefront is a surface on which every point oscillates in phase. Energy travels perpendicular to the wavefront, along the rays. A point source emits spherical wavefronts. Far from the source, a small patch of a huge sphere is practically flat, so a distant source such as the Sun gives a plane wavefront.

Huygens' principle tells you how a wavefront moves. Every point on a wavefront acts as a source of tiny secondary wavelets that spread at the wave speed. After a time t, the new wavefront is the forward envelope, the common tangent, of all those wavelets. The wavelets do not create a backward wave, because their amplitude is largest in the forward direction and zero backwards.

Lenses and prisms reshape wavefronts, and you can predict the result:

Reflection and refraction from Huygens' principle

Let a plane wavefront AB strike an interface obliquely. The edge A touches the boundary first and starts a wavelet in the second medium. While the other edge B travels a distance $BC = v_1 \tau$ to reach the boundary, the wavelet from A spreads to a radius $v_2 \tau$. The new wavefront is the tangent drawn from C to that wavelet. Comparing the two right triangles that share the side AC gives

$$\dfrac{\sin i}{\sin r} = \dfrac{v_1}{v_2} = \dfrac{n_2}{n_1}, \qquad \text{so} \qquad n_1 \sin i = n_2 \sin r$$

This is Snell's law, now derived from waves. It also explains why light bends towards the normal on entering a denser medium: the part of the wavefront that gets in first slows down. For reflection, the wavelets stay in the same medium with the same speed, the triangles are congruent, and the angle of incidence equals the angle of reflection.

The frequency does not change on refraction, because it is fixed by the source and the medium cannot pile up or lose wave crests at the boundary. Since $v = \nu\lambda$, the speed and the wavelength both change by the same factor:

$$\lambda_2 = \lambda_1 \dfrac{v_2}{v_1} = \lambda_1 \dfrac{n_1}{n_2}$$

If the second medium is rarer ($v_2$ is greater than $v_1$), the wave bends away from the normal, and beyond the critical angle $\sin i_c = n_2/n_1$ there is no refracted wave, only total internal reflection.

Worked example: a wave enters water. Yellow light of wavelength 589 nm travels in air and strikes a water surface ($n = 4/3$) at an angle of incidence of 45°. Find its frequency, speed and wavelength in water and the angle of refraction.

Common wrong step. Saying the frequency becomes $3.8 \times 10^{14}$ Hz because "light slows down in water". The source sets the frequency, which is why the colour you see stays the same underwater. The wavelength is what changes.

Interference: when waves meet

When two waves overlap, their displacements add (the principle of superposition). If crest meets crest, the result is bigger. If crest meets trough, the result is smaller. This is interference.

For a steady pattern, the two sources must be coherent: the phase difference between them must stay constant in time. Two sources that start in step and keep that rhythm, like two needles dipping together in a water tank, are coherent. If the phase difference changes randomly and quickly, the bright and dark spots flicker, and the eye just sees an average of uniform brightness.

Let the two waves arriving at a point have equal amplitude $a$ and phase difference $\phi$. The resultant amplitude is $2a\cos(\phi/2)$, so with $I_0$ as the intensity from one source alone,

$$I = 4I_0 \cos^2(\phi/2)$$

The phase and path differences are linked by $\phi = (2\pi/\lambda) \times$ path difference. Note the average intensity of the interference pattern is still $2I_0$. The bright fringes have $4I_0$ and the dark ones $0$, so light is not lost, only redistributed from the dark places to the bright ones.

Two ordinary sodium lamps never give fringes. An ordinary source emits light as short bursts from individual atoms, and its phase changes abruptly about every $10^{-10}$ s. Two independent lamps therefore have a random, rapidly changing phase difference.

Young's double slit experiment

Young solved the coherence problem by using a single narrow slit $S$ to illuminate two close slits $S_1$ and $S_2$. Both then receive the same wave, so any phase jumps happen in both together, and the phase difference between $S_1$ and $S_2$ stays fixed.

Let the slits be $d$ apart, the screen $D$ away, and take a point P at a height $y$ above the central line. For $y$ much smaller than $D$, the path difference is

$$S_2P - S_1P \approx \dfrac{dy}{D}$$

Setting this equal to $n\lambda$ for brightness and $(n + 1/2)\lambda$ for darkness gives

$$\text{Bright: } y_n = \dfrac{n\lambda D}{d}, \qquad \text{Dark: } y_n = \left(n + \dfrac{1}{2}\right)\dfrac{\lambda D}{d}, \qquad \text{Fringe width } \beta = \dfrac{\lambda D}{d}$$

All fringes have the same width, bright and dark are equally spaced, and the centre ($n = 0$, zero path difference) is a bright fringe.

Fringe width depends on the slit separation $d$. It does not depend on how wide each slit is.

Worked example: find the wavelength. In a Young's experiment the slits are 0.28 mm apart and the screen is 1.4 m away. The fourth bright fringe lies 1.2 cm from the central maximum. Find the wavelength of the light.

Common wrong step. Using $y = \lambda D/d$ for the fourth fringe. That is the width of one fringe, not the position of the fourth.

Worked example: change the setup. In a Young's experiment $\lambda = 600 \text{ nm}$, $d = 1.5 \text{ mm}$ and $D = 1.2 \text{ m}$. (a) Find the fringe width. (b) Find it when the apparatus is immersed in water ($n = 4/3$). (c) In air, find the intensity at a point 0.60 mm from the centre, in units of the intensity $I_0$ of one slit.

Common wrong step. Assuming that every point is either bright or dark. At $1.25\lambda$ the point sits between a bright fringe (at $\lambda$) and a dark one (at $1.5\lambda$), and has an intermediate intensity.

Pause and predict: you use white light in a Young's experiment. What do you see at the centre of the screen, and what happens further out?

Check your answer: the centre is white, because at zero path difference every colour has a bright fringe. Moving outwards, each colour's fringes land at slightly different places (violet closest, red farthest), so the pattern turns into coloured bands that soon overlap into a washed out white.

Beyond the textbook. For the exam: $S_1$ and $S_2$ are coherent because they are lit by the same source $S$. In reality: light from a source is coherent over a very short distance in time and space, so the slit $S$ has to be narrow and the slits close together to get clear fringes with an ordinary lamp. A laser's light stays in phase much longer, so with a laser the double slit experiment works easily without S. Why it matters: it explains why fringes fade when the slits are too far apart or the path difference gets large.

Diffraction: waves do not stop at the edge

Light passing through a narrow slit does not simply form a sharp bright strip. It spreads into the geometric shadow and makes a pattern: a wide, bright central maximum with weaker bright bands on either side. This is diffraction.

For a slit of width a lit by a plane wave, the minima (zero intensity) are at

$$a \sin\theta = n\lambda, \qquad n = \pm 1, \pm 2, \pm 3, \ldots \quad (\text{so } \theta \approx n\lambda/a)$$

Why? At the first minimum, $a \sin\theta = \lambda$. Split the slit into two halves. A wavelet from a point in the top half has a partner $a/2$ away in the bottom half with a path difference of $\lambda/2$, so the two cancel exactly. Every point in the slit is cancelled by a partner, so the intensity is zero. At $a \sin\theta = 3\lambda/2$, the slit divides into three parts, two cancel and the third is left over, so there is a weak maximum. The higher secondary maxima are weaker and weaker.

The central maximum extends from $\theta = -\lambda/a$ to $+\lambda/a$:

$$\text{Angular width of central maximum} = \dfrac{2\lambda}{a}, \qquad \text{linear width on a screen at distance } D = \dfrac{2\lambda D}{a}$$

The central maximum is twice as wide as any other bright band, which stretch from $n\lambda/a$ to $(n + 1)\lambda/a$ (width $\lambda/a$). A narrower slit gives a wider pattern. If $a$ is much larger than $\lambda$, the spreading is tiny and light appears to move in straight lines, which is the ray optics of the previous chapter.

Strictly, the double slit pattern is the interference pattern modulated by the diffraction of each slit. There is no sharp physical boundary between interference and diffraction. The names are used for a few interfering sources and for a continuous distribution of sources respectively. In both, the total energy is conserved.

Worked example: slit width and the central maximum. A single slit of width 0.10 mm is illuminated by monochromatic light and a screen 50 cm away shows a central maximum 5.0 mm wide. Find the wavelength, and then the new width if the slit is narrowed to 0.05 mm.

Common wrong step. Using $\lambda D/a$ as the width of the central maximum. That is the distance from the centre to the first minimum, half the full width, and it would give a wavelength of 1000 nm.

Pause and predict: you narrow a slit in front of a laser pointer. Does the spot on the wall get narrower or wider?

Check your answer: wider. The central maximum has width $2\lambda D/a$, so a smaller $a$ makes the pattern spread more. The wave is squeezed in one direction and spreads out in that direction.

Beyond the textbook. For the exam: diffraction needs a slit or obstacle comparable to the wavelength. In reality: the shimmering rainbow on the back of a CD comes from diffraction. The closely spaced tracks act as a grating, and each colour is diffracted at a different angle, so you see the colours separated. Why it matters: it shows the same wave ideas on a small scale, and the spacing of tracks (about 1.6 μm) is comparable to the wavelength of light, which is exactly the condition for visible diffraction.

Polarisation: the proof that light is transverse

In a transverse wave the displacement is perpendicular to the direction of travel. If it always lies in one fixed plane, the wave is linearly (plane) polarised. In ordinary light, such as sunlight or a bulb, the plane of vibration changes randomly and rapidly, so the light is unpolarised. Only transverse waves can be polarised. Sound in air cannot, because its vibrations are along the direction of travel.

A polaroid is a sheet with long, aligned molecules. It absorbs the electric field component along those molecules and transmits the component perpendicular to them, along the pass axis. Unpolarised light passing through one polaroid emerges plane polarised with half the intensity, whatever the polaroid's orientation. If the light is already polarised, its intensity after the polaroid depends on the angle $\theta$ between the polarisation direction and the pass axis. This is Malus' law:

$$I = I_0 \cos^2\theta$$

The field component along the pass axis is $E_0 \cos\theta$, and intensity goes as the square of the field. At $\theta = 0$ everything passes, at 90° (crossed polaroids) nothing does.

Put a third polaroid between two crossed ones and some light reappears. If $I_0$ is the intensity of the light leaving the first polaroid and the middle one is at angle $\theta$ to the first,

$$I = I_0 \cos^2\theta \, \cos^2(90^\circ - \theta) = \dfrac{I_0}{4} \sin^2 2\theta$$

which is largest when $\theta = 45^\circ$.

Worked example: three polaroids. Unpolarised light of intensity $I_0$ passes through three polaroids. The second has its axis at 60° to the first, and the third is crossed with the first. Find the final intensity.

Common wrong step. Applying Malus' law to the unpolarised light at the first polaroid and writing $I_0 \cos^2\theta$. Unpolarised light has no single plane, so the first polaroid always gives half.

Pause and predict: you look at the blue sky through a polaroid and rotate it. Does the brightness change?

Check your answer: yes, a little. Light scattered by air molecules is partly polarised, especially at right angles to the Sun's direction, so a rotating polaroid dims and brightens the sky.

Beyond the textbook. For the exam: polaroids are used in sunglasses, windows and cameras. In reality: sunlight reflected from a road or a lake is partly polarised in the horizontal plane, and polarising sunglasses have their pass axis vertical, so they block much of that glare while letting other light through. The screen of a liquid crystal display also relies on polarising filters to switch pixels on and off. Why it matters: polarisation is not a lab curiosity, it is behind everyday objects.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Previous: Ray Optics and Optical Instruments is the limit of this chapter when wavelength is tiny compared with the apparatus.
• Next: Dual Nature of Radiation and Matter returns to the particle versus wave question and shows light needs both pictures.
• Foundations: Electromagnetic Waves shows that light is a transverse EM wave, which is why it can be polarised.
• Class XI link: Waves introduces superposition and interference of waves on strings and in air, the same ideas used here.