Wave Optics

The chapter has four connected movements. First, Huygens' principle: the definition of a wavefront, the geometrical construction of secondary wavelets, and its use to derive the laws of reflection and refraction — recovering Snell's law and the critical angle from pure wave geometry, and settling the historic dispute between Newton's corpuscular model and the wave model in favour of waves. Second, coherent and incoherent superposition: what it means for two sources to be coherent, why an ordinary pair of lamps never shows fringes, and the general intensity formula $I = 4I_0\cos^2(\phi/2)$ that governs every interference pattern in the chapter. Third, Young's double-slit experiment: the classic demonstration of interference, the path-difference argument that fixes bright and dark fringe positions, and the resulting fringe-width formula. Fourth, diffraction and polarisation: the single-slit diffraction pattern with its central and secondary maxima, and the polarisation of light by a polaroid, culminating in Malus' law and the proof — unavailable to sound waves — that light is transverse.

Wavefronts and Huygens' principle

Drop a stone into still water and every point on the surface it disturbs starts oscillating; a snapshot at any instant shows expanding circles on which the disturbance is at the same stage of oscillation. Such a locus of points vibrating in phase is called a wavefront — a surface of constant phase — and it is the natural generalisation of a "ray" to wave language. A point source emits spherical wavefronts; far from the source, a small patch of a spherical wavefront looks flat, and is treated as a plane wavefront. Energy always flows perpendicular to the wavefront, which is exactly the direction a ray occupies in geometrical optics — a ray is nothing but the trajectory perpendicular to successive wavefronts, in the limit where wavelength is neglected. Huygens' principle is a rule for building the wavefront at a later time from the wavefront now. It states that every point on an existing wavefront acts as a source of new secondary wavelets, spreading out in all directions at the wave's own speed; the surface tangent to all these wavelets in the forward direction — their common envelope — is the new wavefront an instant later. Huygens had to assume, without full justification, that the wavelets have zero amplitude backward and maximum amplitude forward, to suppress an unphysical "backwave"; a fully rigorous justification needs the more advanced Huygens–Fresnel principle, which also underlies diffraction.

Reflection and refraction from Huygens' construction

Applying Huygens' construction where a plane wavefront meets a plane interface separating two media of wave speeds $v_1$ and $v_2$ produces the familiar laws of geometrical optics, but now with a wave-based explanation of why they hold. The refractive indices are defined through the speed of light:

where: $c$ is the speed of light in vacuum and $v_1, v_2$ are the speeds of light in media 1 and 2. A larger refractive index therefore means a slower wave speed — the physical content behind the number that ray optics simply looked up in a table. Working through the geometry (carried out in full in the first derivation block below) gives Snell's law of refraction:

where: $i$ is the angle of incidence, $r$ is the angle of refraction, and both are measured from the normal to the interface. The crucial physical payoff is the direction of bending: if $r < i$ (the ray bends toward the normal, as when light enters glass from air), Huygens' wave picture demands $v_2 < v_1$ — light must travel slower in the denser medium. This is the opposite of what Newton's corpuscular model predicted, and it was this very prediction that Foucault's 1850 experiment confirmed, settling the historic wave-versus-particle dispute in favour of the wave theory. A parallel argument for reflection at a plane mirror (congruent triangles rather than a ratio of speeds) recovers the familiar law $i = r$. A companion result concerns wavelength. If the path $BC$ in the first medium equals one wavelength $\lambda_1$, the corresponding path $AE$ in the second medium equals $\lambda_2$, since both waves take the same time $\tau$ to advance by one full wave. This gives:

where: $\lambda_1, \lambda_2$ are the wavelengths in the two media. On entering a denser medium ($v_2 < v_1$), both speed and wavelength shrink in the same proportion, but the frequency $\nu = v/\lambda$ stays exactly the same — frequency is fixed by the source, not the medium, because the atoms at the boundary re-radiate at the frequency that is driving them. When light travels from a denser medium (1) toward a rarer one (2), so that $v_2 > v_1$, the refracted ray bends away from the normal and $r$ grows faster than $i$. There exists a special angle of incidence, the critical angle $i_c$, at which $r$ reaches exactly $90°$:

where: $n_1 > n_2$ (medium 1 is denser). For any angle of incidence greater than $i_c$, Snell's law would demand $\sin r > 1$, which is impossible — no refracted ray exists, and all the light reflects back into the denser medium. This is total internal reflection, the physics behind optical fibres and the sparkle of a cut diamond, and it falls straight out of the same wavefront construction used for ordinary refraction.

Coherent superposition and interference intensity

Interference is what happens when two or more waves overlap at the same point in space, and the mathematics behind it is the superposition principle: the resultant displacement at any point is simply the vector (here, scalar, since both waves have the same direction of vibration) sum of the displacements each wave would produce alone. Two sources are called coherent if the phase difference between the waves they emit, measured at any fixed point, stays constant in time; light from a single source split into two paths (as in Young's experiment) is coherent, while light from two independent lamps is not, because ordinary sources undergo random phase jumps roughly every $10^{-10}\ \text{s}$. If two coherent waves of the same frequency and amplitude $a$ arrive at a point with phase difference $\phi$, writing $y_1 = a\cos\omega t$ and $y_2 = a\cos(\omega t + \phi)$ and adding them gives a resultant amplitude $2a\cos(\phi/2)$. Since intensity is proportional to amplitude squared, the resultant intensity is:

where: $I_0$ is the intensity each source would produce alone, and $\phi$ is the phase difference at that point. The physical reading is a smooth oscillation between two extremes: at $\phi = 0, \pm 2\pi, \pm 4\pi, \ldots$ (path difference an integer number of wavelengths) the waves arrive crest-on-crest and $I = 4I_0$, constructive interference; at $\phi = \pm\pi, \pm 3\pi, \ldots$ (path difference a half-integer number of wavelengths) the waves arrive crest-on-trough and $I = 0$, destructive interference. If the two sources are incoherent, $\phi$ itself fluctuates randomly and the time-averaged intensity settles to the ordinary sum $I = 2I_0$ everywhere — no bright or dark bands, just uniform illumination. This single formula, Equation (5), is the engine behind every fringe pattern in this chapter.

Young's double-slit experiment and fringe width

Thomas Young's genius was to "lock" two sources in phase by deriving both from a single original source: light from a narrow slit $S$ falls on two closely spaced slits $S_1$ and $S_2$, which then act as coherent secondary sources because any phase change at $S$ affects both identically. The spherical waves spreading from $S_1$ and $S_2$ overlap on a distant screen and produce a pattern of equally spaced bright and dark fringes. For a point $P$ on the screen at distance $x$ from the centre, with slit separation $d$ and screen distance $D \gg d$, the path difference works out (derived below) to $S_2P - S_1P \approx xd/D$. Bright fringes occur where this path difference is a whole number of wavelengths:

where: $x_n$ is the position of the $n$-th bright fringe from the centre, $\lambda$ is the wavelength, $D$ is the slit-to-screen distance and $d$ is the slit separation. Dark fringes occur midway, where the path difference is a half-integer number of wavelengths:

Because consecutive bright fringes (or consecutive dark fringes) are separated by exactly the same distance, the pattern is a set of equally spaced bands, and this spacing — the fringe width — is $\beta = \lambda D/d$, derived formally in the second derivation block below. A smaller slit separation, a longer wavelength, or a more distant screen all widen the fringes; a wider slit separation squeezes them together.

Diffraction at a single slit

Even a single narrow slit, illuminated by a parallel beam of monochromatic light, produces its own pattern of bright and dark bands rather than a sharp-edged shadow. Treating every point across the width of the slit as a Huygens secondary source and summing their contributions at the screen with the correct phase, the intensity has a broad central maximum straight ahead ($\theta = 0$), flanked by much weaker secondary maxima that fade rapidly with distance from the centre. The dark bands — the minima of the pattern — occur at angles satisfying:

where: $a$ is the slit width, $\theta$ is the angle measured from the straight-through direction, and $n$ is a nonzero integer. Between consecutive minima lie the secondary maxima, located approximately at $a\sin\theta \approx (n + \tfrac{1}{2})\lambda$; unlike Young's fringes, these secondary maxima are not evenly spaced in intensity — each is markedly weaker than the last, and the very first secondary maximum carries only a small fraction of the central maximum's intensity. The central maximum is twice as wide as any secondary maximum, spanning from $\theta = -\lambda/a$ to $\theta = +\lambda/a$, because it is bounded by the $n = \pm 1$ minima on either side of the undeviated direction, whereas each secondary maximum is squeezed between consecutive same-sign minima one $\lambda/a$ apart. Diffraction and interference share the same physical root — the superposition of secondary wavelets — and Richard Feynman's famous remark applies precisely here: there is no sharp physical boundary between the two words, only a convention that "interference" is used for a handful of discrete coherent sources (as in Young's experiment) and "diffraction" for a continuum of them (as across a single slit). In fact the double-slit pattern seen on a screen is really the product of the single-slit diffraction envelope from each slit and the Young interference fringes between the two slits — the two effects always coexist.

Polarisation and Malus' law

Everything discussed so far — reflection, refraction, interference, diffraction — works equally well for longitudinal waves like sound. Polarisation does not, and that makes it the decisive proof that light is a transverse wave: the oscillation (the electric field, for light) is always perpendicular to the direction the wave travels. Light from an ordinary source is unpolarised — the direction of the transverse oscillation changes randomly and rapidly, exploring every direction in the plane perpendicular to propagation. A polaroid sheet, built from long-chain molecules aligned in one direction, absorbs the field component along that direction and transmits only the perpendicular component, called the polaroid's pass-axis; the emerging light is linearly (plane) polarised, oscillating along a single fixed direction. Passing unpolarised light through a single ideal polaroid always halves its intensity, regardless of the polaroid's orientation, because on average exactly half of a randomly oriented field survives projection onto any one fixed axis. The interesting physics appears with a second polaroid. If the already-polarised light (amplitude $E$, along the first polaroid's pass-axis) meets a second polaroid whose pass-axis is tilted at angle $\theta$ to the first, only the component $E\cos\theta$ passes through, and since intensity scales as amplitude squared:

where: $I_0$ is the intensity of the light incident on the second polaroid (already polarised by the first) and $\theta$ is the angle between the two pass-axes. This is Malus' law. It predicts full transmission when the axes are parallel ($\theta = 0$), and exactly zero transmission when they are perpendicular — crossed polaroids — at $\theta = 90°$. Rotating one polaroid relative to the other therefore acts as a continuously variable light valve, the working principle behind polarised sunglasses, camera filters, and the intensity control found in many LCD-based displays.

Board level - reflected and refracted wavelength, frequency and speed at an air-water interface

Monochromatic light of wavelength $589\ \text{nm}$ is incident from air on a water surface. Refractive index of water is $n = 1.33$ and $c = 3.0\times10^{8}\ \text{m/s}$. Find the wavelength, frequency and speed of (a) the reflected light and (b) the refracted light.

Step 1. Find the incident frequency, which does not change on reflection or refraction. $\nu = \dfrac{c}{\lambda_1} = \dfrac{3.0\times10^{8}}{589\times10^{-9}} = 5.09\times10^{14}\ \text{Hz}$.

Step 2. Handle the reflected ray. Reflection sends the wave back into the same medium (air), so nothing about the wave changes: wavelength, speed and frequency are all identical to the incident values. Reflected light: $\lambda = 589\ \text{nm}$, $v = 3.0\times10^{8}\ \text{m/s}$, $\nu = 5.09\times10^{14}\ \text{Hz}$.

Step 3. Handle the refracted ray's speed and wavelength. The frequency stays fixed at $5.09\times10^{14}\ \text{Hz}$ (it is set by the source, not the medium), but the speed drops to $v_2 = c/n = (3.0\times10^{8})/1.33 = 2.26\times10^{8}\ \text{m/s}$, and correspondingly the wavelength shrinks to $\lambda_2 = \lambda_1/n = 589/1.33 = 443\ \text{nm}$ (equivalently $\lambda_2 = v_2/\nu$, same answer). Refracted light: $\lambda = 443\ \text{nm}$, $v = 2.26\times10^{8}\ \text{m/s}$, $\nu = 5.09\times10^{14}\ \text{Hz}$ (unchanged).

Board level - fringe wavelength and fringe width in Young's double-slit experiment

In a Young's double-slit experiment the slits are separated by $d = 0.28\ \text{mm}$ and the screen is placed $D = 1.4\ \text{m}$ away. The distance between the central bright fringe and the fourth bright fringe is measured to be $1.2\ \text{cm}$. Determine the wavelength of light used, and the fringe width.

Step 1. Identify which formula applies. The fourth bright fringe sits at $x_4 = \dfrac{4\lambda D}{d}$ from the centre, by Equation (6) with $n = 4$.

Step 2. Solve for $\lambda$. Rearranging, $\lambda = \dfrac{x_4\, d}{4D} = \dfrac{(1.2\times10^{-2}\ \text{m})(0.28\times10^{-3}\ \text{m})}{4 \times 1.4\ \text{m}} = \dfrac{3.36\times10^{-6}}{5.6} = 6.0\times10^{-7}\ \text{m}$. Answer: $\lambda = 600\ \text{nm}$.

Step 3. Find the fringe width directly. $\beta = \dfrac{\lambda D}{d} = \dfrac{(6.0\times10^{-7})(1.4)}{0.28\times10^{-3}} = 3.0\times10^{-3}\ \text{m}$. Answer: $\beta = 3.0\ \text{mm}$, consistent with the given data since four fringe widths span $1.2\ \text{cm}$, i.e. $4 \times 3.0\ \text{mm} = 1.2\ \text{cm}$.

Competitive level (JEE / NEET) - single-slit diffraction: central maximum width and the first secondary maximum

A parallel beam of light of wavelength $\lambda = 600\ \text{nm}$ is incident normally on a single slit of width $a = 0.20\ \text{mm}$. A screen is placed $D = 2.0\ \text{m}$ away. Find (a) the angular half-width and the total linear width of the central maximum, and (b) the angular position of the first secondary maximum.

Step 1. Locate the first minima using Equation (8) with $n = \pm 1$: $\sin\theta_1 = \dfrac{\lambda}{a} = \dfrac{6.0\times10^{-7}}{2.0\times10^{-4}} = 3.0\times10^{-3}$. Since this is small, $\theta_1 \approx 3.0\times10^{-3}\ \text{rad}$ (in radians, using the small-angle approximation).

Step 2. Convert to a linear half-width on the screen: $y_1 = D\theta_1 = 2.0 \times 3.0\times10^{-3} = 6.0\times10^{-3}\ \text{m} = 6.0\ \text{mm}$. The central maximum spans from $-y_1$ to $+y_1$, so its total width is $12\ \text{mm}$, twice the half-width — matching the general rule that the central maximum is twice as wide as any secondary maximum.

Step 3. Locate the first secondary maximum using the approximate secondary-maximum condition $a\sin\theta \approx (n+\tfrac12)\lambda$ with $n = 1$: $\sin\theta' = \dfrac{1.5\lambda}{a} = 1.5 \times 3.0\times10^{-3} = 4.5\times10^{-3}$, so $\theta' \approx 4.5\times10^{-3}\ \text{rad}$, landing at $y' = D\theta' = 9.0\ \text{mm}$ from the centre — well outside the central maximum's $6.0\ \text{mm}$ half-width, as it must be. Answer: central maximum half-width $\approx 3.0\times10^{-3}\ \text{rad}$ (total width $12\ \text{mm}$); first secondary maximum at $\approx 4.5\times10^{-3}\ \text{rad}$ ($9.0\ \text{mm}$ from centre). Timing note: under two minutes if you keep the pattern straight — minima sit at integer multiples of $\lambda/a$, the central maximum's half-width is exactly the first minimum's angle, and secondary maxima sit roughly at half-integer multiples, always strictly between two consecutive minima. The classic JEE trap is plugging $n=1$ into the minima formula $a\sin\theta = n\lambda$ when the question actually asks for a secondary maximum — that gives the second minimum's angle by mistake, not the first bright band beside the centre. Always check whether the question wants a dark band (integer $n$) or a bright band (half-integer $n$) before substituting.

Students consistently mix up a handful of specific ideas in this chapter, and examiners know exactly where to press.

Adding intensities instead of amplitudes for coherent sources. The instinct from "two bulbs are brighter than one" is to write $I = I_1 + I_2$ always. That rule holds only for incoherent sources; for coherent sources the correct general formula is Equation (5), $I = 4I_0\cos^2(\phi/2)$, which swings all the way from $4I_0$ down to zero depending on phase — a fact ray-optics intuition about brightness never anticipates.

Forgetting that frequency, not wavelength, is what stays fixed across a boundary. On refraction, both speed and wavelength change together (Equation 3), but frequency is locked to the source and is identical in both media — a point tested directly in NCERT's own Example 10.1(a).

Confusing path difference with phase difference. A path difference of $\Delta$ corresponds to a phase difference $\phi = (2\pi/\lambda)\Delta$; forgetting the $2\pi/\lambda$ factor when converting between the two is a very common slip, especially in problems that quote path difference in multiples of $\lambda$ rather than in metres.

Mixing up the single-slit minima condition with the double-slit maxima condition. Both look like "$(\text{something})\sin\theta = n\lambda$," but in Young's experiment $n\lambda$ marks a bright fringe (Equation 6) while in single-slit diffraction $n\lambda$ marks a dark minimum (Equation 8) — memorising the formula without the physical picture behind it (constructive addition of a handful of point sources versus destructive cancellation across a continuum of wavelets) leads straight to sign and interpretation errors.

Applying Malus' law with the wrong reference intensity. $I = I_0\cos^2\theta$ uses $I_0$ as the intensity of the light already polarised and incident on the polaroid in question, not the original unpolarised intensity — skip the "divide by two for the first polaroid" step and every downstream number in a multi-polaroid problem comes out wrong, exactly the trap probed in the three-polaroid derivation above.

Believing total internal reflection needs going "from air to glass." It only occurs going from a denser to a rarer medium at an angle beyond the critical angle $i_c = \sin^{-1}(n_2/n_1)$ — light travelling from air into glass can never totally internally reflect, no matter the angle, because it is entering the denser medium.

• State Huygens' principle precisely and use it to sketch the construction of a refracted or reflected wavefront at a plane interface.
• Derive Snell's law and the critical angle condition from Huygens' construction, and explain why the wave model's prediction about speed in a denser medium contradicts the corpuscular model's.
• Distinguish coherent from incoherent sources, and explain physically why two independent lamps never produce a stable interference pattern.
• Apply the general interference intensity formula $I = 4I_0\cos^2(\phi/2)$ to identify points of constructive and destructive interference from a phase or path difference.
• Derive the fringe-width formula $\beta = \lambda D/d$ for Young's double-slit experiment and use it (in either direction) to find wavelength from measured fringe spacing or vice versa.
• Locate the minima and estimate the secondary maxima of a single-slit diffraction pattern using $a\sin\theta = n\lambda$, and explain why the central maximum is twice as wide as any secondary maximum.
• State Malus' law, apply it correctly across a sequence of polaroids, and explain the transverse-wave evidence it provides.
• Explain, using conservation of energy, why interference and diffraction only redistribute light rather than creating or destroying it.

Ask the tutor: If you want to see exactly how the fringe-width formula changes when the double slit is dipped in water, or work through why a diffraction grating gives much sharper lines than a single double slit, ask your tutor to walk through it with you.