Wave Optics

In Chapter 9 we treated light as rays travelling in straight lines. That works because the wavelength of light (about 4 × 10⁻⁷ m to 7 × 10⁻⁷ m) is far smaller than the mirrors, lenses and apertures we normally use. This chapter is about the situations where the wave nature shows itself: light passing through very narrow slits, two beams overlapping, light bending round an edge.

A short history is expected in the introduction. Descartes (1637) explained refraction with a corpuscular model, which Newton developed. In that model, light corpuscles are pulled toward a denser medium at the boundary, so the normal component of velocity increases and the ray bends toward the normal. This predicts that light travels faster in a denser medium. Huygens (1678) proposed the wave theory, which explains reflection and refraction equally well but predicts that light travels slower in a denser medium. Foucault (1850) measured the speed of light in water and found it lower than in air, which settled the matter in favour of the wave theory. Before that, Young's interference experiment (1801) had already shown that light is a wave, because interference and diffraction cannot be explained by particles travelling in straight lines.

Maxwell (1864) showed that light is an electromagnetic wave whose speed in vacuum is c = 1/√(μ₀ε₀), and Hertz produced such waves in 1887. Light waves are transverse: the electric and magnetic fields oscillate perpendicular to the direction of travel. That transverse nature is what polarisation (Section 10.7) demonstrates.

The plan of the chapter: Huygens' principle first, then its use to prove the laws of reflection and refraction, then the superposition of two waves (interference), the bending of a wave passing a single slit (diffraction), and finally polarisation. Keep in mind throughout that when light crosses from one medium to another, its frequency does not change; the speed and wavelength change in the same ratio.

A wavefront is the locus of all points in the medium that are vibrating in the same phase. For a point source in a uniform medium the wavefronts are concentric spheres (spherical wavefront); at a large distance from the source a small portion of a sphere is nearly flat, so we speak of a plane wavefront; a line source gives cylindrical wavefronts. A ray is a line drawn perpendicular to the wavefront in the direction of propagation. Energy travels along the rays, perpendicular to the wavefront.

Huygens' principle tells us how to find the wavefront at a later time if we know it now. Each point on a wavefront acts as a source of a secondary disturbance, and the wavelets from these points spread out in all directions with the speed of the wave. The new wavefront at a time τ later is the common tangent (envelope) to all these secondary wavelets, drawn in the forward direction.

Construction, as the examiner wants it: take a spherical wavefront F₁F₂ at t = 0 from a source O. Mark points A, B, C, D on it. With each as centre draw a sphere of radius vτ, where v is the speed of the wave. The common tangent G₁G₂ on the forward side is the wavefront at time τ; it is again a sphere with centre O, of radius larger by vτ. For a plane wavefront the same construction gives a parallel plane wavefront.

One point deserves a sentence in a 3-mark answer. The wavelets also have a common tangent on the backward side, which would predict a backward-moving wave. Huygens argued that the amplitude of the secondary wavelets is zero in the backward direction, so there is no back wave. This is an assumption of the model, not a derived result.

Standard 2-mark asks about shapes: a plane wavefront on a convex lens emerges as a spherical wavefront converging on the focus; on a concave mirror it reflects as a converging spherical wavefront; on a prism it emerges as a tilted plane wavefront (the part passing through the thicker side is delayed more); from a point source at the focus of a convex lens it emerges as a plane wavefront.

This is the derivation the board asks most often from this chapter. Set it up exactly like NCERT Fig. 10.4. A plane wavefront AB is incident on the surface PP' separating medium 1 (speed v₁) from medium 2 (speed v₂ < v₁). The incident ray makes angle i with the normal, so the wavefront AB makes angle i with the surface. Let τ be the time in which the disturbance from B reaches the surface at C. Then BC = v₁τ.

In the same time τ the secondary wavelet from A has travelled a distance AE = v₂τ into medium 2. Draw a sphere of radius v₂τ about A; the tangent from C to this sphere touches it at E. CE is the refracted wavefront, and it makes angle r with the surface, so the refracted ray makes angle r with the normal.

From triangle ABC, sin i = BC/AC = v₁τ/AC. From triangle AEC, sin r = AE/AC = v₂τ/AC. Dividing:

sin i / sin r = v₁/v₂

This ratio is a constant for the pair of media, which is Snell's law. If c is the speed in vacuum, n₁ = c/v₁ and n₂ = c/v₂ are the refractive indices, and the law becomes n₁ sin i = n₂ sin r. Also, the incident ray, the refracted ray and the normal all lie in the plane of the figure, which is the other part of the law of refraction.

Two consequences that fetch the last mark. First, since v₂ < v₁, sin r < sin i, so r < i: the ray bends toward the normal in the denser medium, in agreement with experiment and against the corpuscular model. Second, the frequency does not change on refraction, because the number of waves arriving at the surface per second must equal the number leaving it. So if λ₁ and λ₂ are the wavelengths, v₁ = νλ₁ and v₂ = νλ₂, giving λ₁/λ₂ = v₁/v₂ = n₂/n₁: the wavelength is shorter in the denser medium.

Now let the wave go from a denser medium into a rarer one, so v₂ > v₁. The same construction applies: BC = v₁τ and AE = v₂τ, but now AE is larger than BC, and the refracted wavefront CE is tilted further away. Snell's law still reads n₁ sin i = n₂ sin r, and since n₂ < n₁ we get sin r > sin i: the refracted ray bends away from the normal.

Because r must be larger than i, there is a limit. Define the critical angle i_c as the angle of incidence for which r = 90°. Then n₁ sin i_c = n₂ sin 90°, so sin i_c = n₂/n₁ = n₂₁, where n₂₁ < 1 for a rarer second medium. For i > i_c, sin r would have to exceed 1, which is impossible, so no refracted wave can be constructed: the sphere of radius v₂τ about A is now so large that no tangent from C can be drawn to it. The wave is entirely reflected. This is total internal reflection, seen here from the wave point of view.

The link to Chapter 9 is worth one line in an answer: for glass with n = 1.5 in air, sin i_c = 1/1.5, i_c ≈ 42°, which is why a 45° prism turns light through 90° or 180° by total internal reflection.

A short numerical the board likes: light of wavelength 600 nm in air enters glass of refractive index 1.5. The speed becomes 2 × 10⁸ m s⁻¹, the wavelength becomes 400 nm, and the frequency stays 5 × 10¹⁴ Hz. Coming out of the glass, all three return to their air values.

A plane wavefront AB is incident on a reflecting surface MN at an angle of incidence i, with v the speed of the wave in the medium. As before, let τ be the time in which the disturbance from B reaches the surface at C, so BC = vτ. In the same time the secondary wavelet from A has spread out to a sphere of radius AE = vτ. Draw the tangent from C to this sphere; it touches at E. CE is the reflected wavefront, and the reflected ray makes angle r with the normal.

Compare the triangles EAC and BAC. AE = BC = vτ, AC is common, and both are right-angled (angle AEC and angle ABC are 90° because a wavefront is perpendicular to the ray). So the two triangles are congruent, and the angle BAC equals the angle ECA, that is, i = r. The incident ray, the reflected ray and the normal lie in the same plane. This proves the law of reflection.

The same method extends to lenses, mirrors and prisms; what you need for the exam are the resulting wavefront shapes. A plane wavefront through a thin prism emerges as a plane wavefront tilted toward the base, because the part through the thicker portion is slowed for longer. A plane wavefront through a convex lens emerges as a spherical wavefront converging on the focus, since the central part passes through the thickest glass and is delayed most. A plane wavefront on a concave mirror reflects as a spherical wavefront converging on the focus. In every case the total time taken by light from a point on the object to the corresponding point on the image is the same along every ray, which is why the rays can converge to form an image.

Reflected light keeps the same speed, wavelength and frequency as the incident light because it stays in the same medium.

Superposition says that at a point where two waves overlap, the resultant displacement is the sum of the individual displacements. Interference is the effect of this sum on the intensity. Take two needles S₁ and S₂ moving up and down in phase on a water surface; they are coherent sources because the phase difference between them does not change with time.

At a point P equidistant from both, the two waves arrive in phase: y₁ = y₂ = a cos ωt, so y = 2a cos ωt and, since intensity is proportional to the square of the amplitude, I = 4I₀, where I₀ is the intensity due to one source alone. At a point Q where S₂Q − S₁Q = 2λ, the waves arrive with phase difference 4π, again in phase, again I = 4I₀. At a point R where S₂R − S₁R = −2.5λ, the phase difference is 5π, the waves are exactly out of phase, y = 0 and I = 0.

In general, if S₂P − S₁P = nλ (n = 0, 1, 2, ...) the waves arrive in phase and the intensity is 4I₀: constructive interference. If S₂P − S₁P = (n + ½)λ, the intensity is zero: destructive interference. For any phase difference φ, write y₁ = a cos ωt, y₂ = a cos(ωt + φ). Then y = y₁ + y₂ = 2a cos(φ/2) cos(ωt + φ/2). The amplitude is 2a cos(φ/2), so

I = 4I₀ cos²(φ/2)

which gives 4I₀ for φ = 0, ±2π, ±4π, ... and 0 for φ = ±π, ±3π, .... For unequal amplitudes a₁ and a₂, the resultant amplitude squared is a₁² + a₂² + 2a₁a₂ cos φ, so I_max/I_min = (a₁ + a₂)²/(a₁ − a₂)².

Now the key point. If the two sources are incoherent, φ changes randomly with time, so cos²(φ/2) averages to ½ and I = 2I₀ everywhere: the intensities simply add and no pattern is seen. Two independent light sources, even two identical sodium lamps, are incoherent because each atom emits a wave train for only about 10⁻¹⁰ s and the phase of the emitted light changes randomly. That is why Young had to derive both his sources from one.

Energy is not destroyed in interference; it is redistributed. Averaged over a fringe, the intensity is 2I₀, exactly the sum of the two.

Young (1801) illuminated a single pinhole S and then two pinholes S₁ and S₂ very close together, equidistant from S. Light from S reaches both with the same phase, so S₁ and S₂ act as coherent sources. On a screen GG' at a distance D they produce alternate bright and dark bands: the interference fringes.

Positions of the fringes. Let the slits be d apart and P a point on the screen at distance x from the central point O (which is equidistant from the slits). Then

(S₂P)² − (S₁P)² = [D² + (x + d/2)²] − [D² + (x − d/2)²] = 2xd

so S₂P − S₁P = 2xd/(S₂P + S₁P). Since x and d are both much smaller than D, S₂P + S₁P ≈ 2D, and the path difference is

S₂P − S₁P ≈ xd/D

Bright fringes (constructive interference) occur where xd/D = nλ, that is, at x_n = nλD/d, n = 0, ±1, ±2, .... The n = 0 fringe at O is the central bright fringe. Dark fringes occur at x_n = (n + ½)λD/d. The separation between consecutive bright (or dark) fringes is the fringe width

β = x_n+1 − x_n = λD/d

The fringes are equally spaced and, in this approximation, of equal intensity 4I₀. The angular fringe width is β/D = λ/d, independent of D. Strictly, points of constant path difference lie on hyperbolas, but near the centre with D large these are nearly straight lines parallel to the slits.

What the examiner asks next. (i) β ∝ λ: red fringes are wider than blue. (ii) β ∝ D and β ∝ 1/d: moving the screen away widens the fringes; moving the slits apart narrows them; the angular width λ/d does not change with D. (iii) In a liquid of refractive index n the wavelength becomes λ/n, so β becomes β/n. (iv) With white light the central fringe is white (zero path difference for every colour) and is flanked by coloured fringes, since each colour has its own fringe width β ∝ λ. NCERT's description: at the point where the path difference is λ_blue/2 the blue component is missing, so the fringe next to the central one looks red; a little farther, where the path difference is λ_blue = λ_red/2, red is missing and the fringe looks blue. After a few fringes the colours overlap and no clear pattern is seen. (v) If one slit is covered, the fringes disappear and only the single-slit diffraction pattern remains. (vi) If the source slit S is widened beyond a limit, each part of it produces a shifted pattern and the fringes blur; the condition for clear fringes is s/S < λ/d, where s is the source-slit width and S its distance from the double slit. (vii) If the source S is moved closer to the double slit, β does not change (it does not involve that distance), but S decreases in the condition s/S < λ/d, so the fringes become less sharp and finally disappear.

Conditions for sustained interference: the two sources must be coherent (constant phase difference, same frequency), preferably of equal amplitude for good contrast, close together, and the screen reasonably far away.

Diffraction is the bending of a wave round the edge of an obstacle or aperture. Sound diffracts visibly round corners because its wavelength is metres; light diffracts noticeably only when the aperture is comparable to its wavelength, so the effect is easy to miss but always present. Since the wavelength of light is small, diffraction effects are small, which is why ray optics usually works.

The single slit. A parallel beam (plane wavefront) of wavelength λ falls on a slit of width a, and the pattern is observed on a screen far away (or at the focal plane of a lens). By Huygens' principle every point of the slit is a source of secondary wavelets. Consider the light leaving at angle θ to the incident direction. Between the two edges of the slit the path difference is a sin θ ≈ aθ for small θ.

At θ = 0 all wavelets are in phase and add to give the central maximum. Now consider θ such that aθ = λ. Divide the slit into two halves. Each point in the upper half has a corresponding point in the lower half a/2 below it, and the path difference between the pair is (a/2)θ = λ/2. The pairs cancel, and the net intensity is zero: this is the first minimum. Dividing into four, six, ... parts shows minima at

a sin θ = nλ, n = ±1, ±2, ±3, ... (θ ≈ nλ/a)

Between these lie secondary maxima at approximately θ ≈ (n + ½)λ/a. At θ = 3λ/2a, for instance, the slit can be divided into three parts of width a/3; two of them cancel in pairs and only the third contributes, so the first secondary maximum is much weaker than the central one (about 1/22 of it). The intensity falls further for higher orders because a smaller fraction of the slit contributes net amplitude.

Width of the central maximum. It extends from θ = −λ/a to θ = +λ/a, so its angular width is 2λ/a and its linear width on a screen at distance D is 2λD/a. Halving the slit width doubles the width of the central maximum and reduces its intensity to one quarter (less light and spread over twice the width). Increasing λ widens the pattern; red light gives a broader pattern than blue.

Double slit as a product. In Young's experiment each slit has a finite width a, so the interference fringes of spacing λD/d sit inside the single-slit diffraction envelope of central width 2λD/a. NCERT Example 10.4: with d = 1 mm, ten interference maxima fit inside the central diffraction maximum if 10 λ/d = 2λ/a, so a = d/5 = 0.2 mm. The two effects are not different in kind: interference is superposition of a few coherent sources, diffraction is superposition of a continuous set of sources across one aperture.

Interference versus diffraction pattern. (i) Interference fringes are all of nearly equal intensity; in diffraction the central maximum is much brighter and the secondary maxima fade quickly. (ii) Interference fringes are equally spaced; the central diffraction maximum is twice as wide as the secondary maxima, which are of equal width. (iii) Interference needs two coherent sources; diffraction comes from a single aperture.

NCERT describes a home experiment worth a 2-mark answer. Hold two razor blades so that their edges are parallel and form a narrow slit, and look through it at a clear glass electric bulb with a straight filament, with the filament parallel to the slit. Put a coloured filter (red or blue) in front of the eye. You will see a bright central band with fainter bands on either side: the single-slit pattern. Without the filter the bands are coloured, because each wavelength gives its own pattern of different width, red wider than blue, and they overlap.

Notice two things. The slit here is the razor-blade gap, the filament plays the role of a distant line source, and the eye lens focuses the parallel diffracted beams onto the retina, which plays the role of the screen at a large distance. Second, the filament must be parallel to the slit: a filament at any other angle acts as many slits at different positions and blurs the pattern.

Diffraction is also why every shadow has a slightly soft edge, and why a bright spot appears at the centre of the shadow of a small circular disc (the wavelets from all points of the edge reach the centre in phase). It is the same effect that limits how fine a detail a telescope or microscope can resolve, since the image of a point is never a point but a small diffraction pattern.

One more remark from NCERT worth knowing: when a slit is wide, diffraction is negligible and the light travels in a straight beam, which is the ray-optics limit. The narrower the slit compared with λ, the more the light spreads. That is the sense in which ray optics is the short-wavelength limit of wave optics.

Polarisation is in the NCERT chapter but has been outside the CBSE syllabus since 2023-24, so read this for completeness and for competitive exams. A wave on a string that moves up and down is linearly (plane) polarised in the vertical plane; a wave whose displacement direction changes randomly is unpolarised. Light is a transverse electromagnetic wave. In ordinary light from a lamp or the Sun, the electric vector points in all directions perpendicular to the ray, changing randomly and rapidly: the light is unpolarised. If the electric vector oscillates along one fixed line perpendicular to the direction of propagation, the light is linearly polarised.

A polaroid is a thin sheet with long-chain molecules aligned in one direction. The electric field component parallel to the chains is absorbed; the component perpendicular to them passes. The perpendicular direction is called the pass-axis. When unpolarised light passes through a polaroid P₁, it emerges linearly polarised along the pass-axis with half the original intensity (the average of cos² over all orientations is ½). If this light falls on a second polaroid P₂ whose pass-axis makes angle θ with that of P₁, only the component of the electric field along the new axis, E₀ cos θ, passes. Since intensity is proportional to the square of the amplitude,

I = I₀ cos² θ (Malus's law)

where I₀ is the intensity of the polarised light incident on P₂. The transmitted intensity is maximum when the axes are parallel (θ = 0) and zero when they are crossed (θ = 90°). Rotating P₂ through one full turn gives two maxima and two minima. That the intensity varies with θ at all is proof that light is a transverse wave; a longitudinal wave such as sound cannot be polarised.

A classic numerical: a third polaroid inserted between two crossed polaroids with its axis at angle θ to the first transmits (I₀/2) cos² θ, and the last one then passes (I₀/2) cos² θ cos²(90° − θ) = (I₀/8) sin² 2θ, which is maximum, I₀/8, at θ = 45°. Polaroids are used in sunglasses, window panes and car windscreens to cut glare, in photographic filters, and in 3-D cinema where the two images are sent with perpendicular polarisations. Polarisation by scattering (the blue of the sky) and by reflection (Brewster's law, tan i_B = n) have been removed from the rationalised book.