NCERT Solutions for Class 12 Physics Chapter 10: Wave Optics
All 6 exercise questions of the latest NCERT textbook (2026–27 reprint) solved like a topper’s answer-sheet — Huygens principle, wavefronts, refraction and speed of light, Young’s double-slit experiment and intensity of interference fringes — with diagrams linked to the exact NCERT pages, comparison tables, complete revision notes (including diffraction and polarisation) and exam tips.
Chapter at a glance — complete revision notes
10.1 Introduction
- Corpuscular model: in 1637 Descartes gave the corpuscular model of light and derived Snell’s law; Newton developed it in his book Opticks. It explains reflection and refraction but predicts that when a ray bends towards the normal the speed of light is greater in the second (denser) medium.
- Wave model: in 1678 Christiaan Huygens proposed the wave theory. It explains reflection and refraction and predicts that if the wave bends towards the normal the speed of light is less in the denser medium — confirmed experimentally by Foucault (1850) (speed in water < speed in air). Newton’s authority and the question “how can waves travel in vacuum?” delayed its acceptance. Thomas Young’s interference experiment (1801) firmly established the wave nature; the wavelength of visible light is extremely small (yellow light ≈ 0.6 μm), so light appears to travel in straight lines (geometrical optics: a ray is the path of energy propagation in the limit λ → 0). The difficulty of propagation through vacuum was removed by Maxwell’s electromagnetic theory of light (c from the wave equation; Hertz produced radio waves around 1890; J. C. Bose and Marconi applied them).
- Plan: Huygens principle and laws of reflection/refraction (10.2–10.3); interference based on the superposition principle (10.4–10.5); diffraction by the Huygens–Fresnel principle (10.6); polarisation — light waves are transverse (10.7).
10.2 Huygens principle
- Wavefront: the locus of points that oscillate in phase — a surface of constant phase; the energy travels perpendicular to the wavefront with the speed of the wave. A point source in a uniform medium gives spherical wavefronts (Fig. 10.1a); at a large distance a small portion is a plane wavefront (Fig. 10.1b).
- Huygens principle: a geometrical construction — every point of a wavefront is a source of secondary wavelets that spread out in all directions with the speed of the wave; the forward envelope (common tangent) of the wavelets at a later time t gives the new wavefront (Fig. 10.2: spheres of radius vt). The ad hoc assumption that the amplitude of the wavelets is maximum forward and zero backward removes the “back wave” (justified properly by the Huygens–Fresnel theory).
10.3 Refraction and reflection of plane waves using Huygens principle
10.3.1 Refraction of a plane wave (Figs 10.3, 10.4)
Plane wavefront AB incident at angle i on the surface PP′ separating media 1 and 2 with speeds v₁ and v₂; during the time t in which B reaches C (BC = v₁t), a sphere of radius v₂t is drawn from A and the tangent CE is the refracted wavefront (AE = v₂t). From the triangles ABC and AEC: sin i / sin r = v₁/v₂. Hence if the wave bends towards the normal (r < i), v₂ < v₁ — opposite to the corpuscular prediction. With n = c/v: n₁ sin i = n₂ sin r (Snell’s law) and λ₁/λ₂ = v₁/v₂ = n₂/n₁: when light enters a denser medium both speed and wavelength decrease but the frequency ν = v/λ remains the same.10.3.2 Refraction at a rarer medium (Fig. 10.5)
For v₂ > v₁ the wave bends away from the normal; the critical angle sin i_c = n₂/n₁; for i > i_c there is no refracted wave — total internal reflection.10.3.3 Reflection of a plane wave (Fig. 10.6)
The reflected wavefront is the tangent plane CE from C to the sphere of radius vt drawn about A; triangles EAC and BAC are congruent so i = r. Prisms, lenses, mirrors (Fig. 10.7): the wavefront through a thin prism tilts because light is slower in glass (the thicker part delays it more); the plane wave through a convex lens becomes spherical and converges to the focus (the central part, travelling through the thickest glass, is delayed most); a concave mirror turns a plane wave into a converging spherical wave. The time taken from a point of the object to the corresponding image point is the same along every ray (the central ray travels a shorter path but is slowed down in the glass).- Example 10.1 / conceptual points: the reflected and refracted light have the same frequency as the incident light because atoms behave as oscillators that take up the frequency of the incident light (forced oscillations); a lower speed in a denser medium does not imply a reduction in the energy carried (energy depends on the amplitude, not on the speed); in the photon picture the intensity is determined by the number of photons crossing unit area per unit time (for a given frequency).
10.4 Coherent and incoherent addition of waves
- Interference rests on the principle of superposition: at a given point the resultant displacement is the vector sum of the displacements due to the individual waves. Two sources are coherent if the phase difference between the displacements they produce at a point does not change with time (e.g. two needles oscillating in phase in a water trough, Fig. 10.8).
- For two coherent sources S₁ and S₂ vibrating in phase, a point P at path difference S₁P ~ S₂P = nλ (n = 0, 1, 2, …) has constructive interference with intensity 4I₀ (displacements y₁ = a cos ωt, y₂ = a cos ωt add to 2a cos ωt; I ∝ amplitude²); at path difference (n + ½)λ the displacements are out of phase and the intensity is zero — destructive interference (Fig. 10.9). At an arbitrary point with phase difference φ the amplitude is 2a cos(φ/2) and the intensity I = 4I₀ cos²(φ/2) (Fig. 10.10): maximum at φ = 0, ±2π, ±4π …, zero at φ = ±π, ±3π …. A path difference λ corresponds to a phase difference 2π.
- If the phase difference changes rapidly with time the sources are incoherent and the intensities simply add (I = 2I₀) — as for two separate light sources illuminating a wall. In interference and diffraction light energy is only redistributed (no gain or loss: conservation of energy).
10.5 Interference of light waves and Young’s experiment
- Two sodium lamps give no fringes (Fig. 10.11) because ordinary sources emit light with abrupt phase changes every ≈ 10⁻¹⁰ s — incoherent. Young’s trick: a pinhole S illuminates two close pinholes S₁ and S₂ (Fig. 10.12), so that S₁ and S₂ derive their light from the same source and are phase-locked — coherent.
- Young’s double-slit: slits separated by d, screen at distance D (D ≫ d): path difference at a point x from the centre ≈ xd/D. Bright fringes: x_n = nλD/d (n = 0, ±1, ±2, …); dark fringes: x = (n + ½)λD/d. Fringe width β = λD/d — equal for bright and dark, evenly spaced (Fig. 10.13, computer-generated fringe pattern with d = 0.025 mm, D = 5 cm, λ = 5 × 10⁻⁵ cm); the fringe spacing ∝ λ, so red fringes are wider than blue; white light gives a central white fringe with coloured fringes. The pattern is actually the double-slit interference pattern superposed on the single-slit diffraction from each slit.
10.6 Diffraction
- Near the geometrical shadow of an opaque object there are alternate dark and bright regions: diffraction — a general property of all waves (sound, light, water, matter waves); light appears to “turn corners”. It is not seen in everyday life because the wavelength of light is much smaller than the dimensions of ordinary obstacles; but the finite resolution of the eye and of telescopes and microscopes is limited by diffraction, and the colours seen on a CD are diffraction effects.
10.6.1 The single slit (Figs 10.14–10.16)
For a slit of width a illuminated by a parallel beam, a broad pattern appears with a central maximum (twice as wide as the others) and weaker secondary maxima on either side. Divide the slit into small parts and add contributions with proper phases (Huygens–Fresnel). Minima (zero intensity) at angles sin θ ≈ θ = nλ/a (n = ±1, ±2, …); secondary maxima at θ ≈ (n + ½)λ/a, getting weaker with increasing n. Half-angular width of the central maximum λ/a: the central maximum spreads with narrower slits. Feynman’s remark: there is no specific, important physical difference between interference and diffraction — “interference” for a few sources, “diffraction” for a large number.10.6.2 Seeing the single slit pattern
Two razor blades held with parallel edges to form a narrow slit, a clear glass bulb with a straight filament: the filament plays the role of the first slit S and the eye lens focuses the pattern on the retina; bands show colours (the positions depend on λ); red and blue filters show the wider fringes for red. A double slit cut in aluminium foil allows a Young’s experiment; never look directly at the Sun (the reflection of the Sun on a shiny convex surface such as a cycle bell can be used).
10.7 Polarisation
- A transverse wave on a string: y(x, t) = a sin(kx − ωt) is a transverse, linearly polarised (plane polarised, y-polarised) wave (Fig. 10.17); a wave whose plane of vibration changes randomly in very short intervals is unpolarised. Light is a transverse electromagnetic wave (E ⟂ direction of propagation) — shown by a polaroid (long chain molecules aligned in one direction absorb the E-component along their length, so the transmitted E oscillates perpendicular to them — the pass-axis).
- Unpolarised light through one polaroid: intensity reduced to half, independent of the rotation; through two polaroids (P₂ then P₁, Fig. 10.18): Malus’ law I = I₀ cos²θ (θ = angle between the pass-axes; E cos θ passes); crossed polaroids (θ = 90°) transmit nothing. Uses: control of intensity, sunglasses, windowpanes, cameras, 3-D displays. Example 10.2: a polaroid rotated between two crossed polaroids: I₃ = (I₀/4) sin² 2θ — maximum at θ = π/4.
Summary & “Points to ponder” (NCERT)
- Huygens’ principle (forward envelope of secondary wavelets) leads to the laws of reflection and refraction and to the equality of travel time along every ray; superposition gives an interference term in addition to the sum of intensities, important only for coherent sources (same frequency, stable phase difference); Young’s double slit gives equally spaced fringes of width λD/d; a single slit of width a gives a central maximum with zeros at nλ/a; natural light is unpolarised, a polaroid transmits only one component, Malus’ law I = I₀cos²θ.
- Points to ponder: waves from a point source spread in all directions while light travels along narrow rays — it took Huygens, Young and Fresnel to explain this; diffraction defines the limits of ray optics (the resolving limit of microscopes and telescopes); most interference and diffraction effects exist even for longitudinal waves like sound, but polarisation is special to transverse waves.
Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light? Refractive index of water is 1.33.
Concept and context. When light is incident on the surface separating two media, the frequency of the reflected and refracted light equals the frequency of the incident light: the atoms of the medium behave as oscillators that are driven at the frequency of the incident wave (forced oscillations), and the frequency of the light they re-emit equals their frequency of oscillation. The speed and wavelength depend on the medium: v = c/n and λ = v/ν = λ₀/n (λ₀ in vacuum/air). Using the wave theory (Huygens’ construction, NCERT Fig. 10.4): sin i/sin r = v₁/v₂ and λ₁/λ₂ = v₁/v₂ = n₂/n₁. The reflected light stays in the first medium (air) and therefore has the same speed, wavelength and frequency as the incident light.
Given: λ = 589 nm in air (n ≈ 1; c = 3.00 × 10⁸ m s⁻¹); n_water = 1.33.
(a) Reflected light (in air)
- Wavelength = 589 nm (the light remains in air).
- Speed = c = 3.00 × 10⁸ m s⁻¹.
- Frequency ν = c/λ = (3.00 × 10⁸)/(589 × 10⁻⁹) = 5.09 × 10¹⁴ Hz.
(b) Refracted light (in water)
- Frequency: unchanged: ν = 5.09 × 10¹⁴ Hz (frequency is a property of the source and does not change on reflection or refraction).
- Speed: v = c/n = (3.00 × 10⁸)/1.33 = 2.26 × 10⁸ m s⁻¹.
- Wavelength: λ′ = v/ν = (2.26 × 10⁸)/(5.09 × 10¹⁴) = 4.44 × 10⁻⁷ m, or λ′ = λ/n = 589/1.33 = 443 nm.
| Quantity | Incident / reflected (air) | Refracted (water) |
|---|---|---|
| Frequency ν | 5.09 × 10¹⁴ Hz | 5.09 × 10¹⁴ Hz (same) |
| Speed v | 3.00 × 10⁸ m s⁻¹ | 2.26 × 10⁸ m s⁻¹ |
| Wavelength λ | 589 nm | 443 nm |
- Colour: since the colour of light is determined by its frequency (not wavelength), the light is still yellow in water, although its wavelength has dropped to 443 nm (which would be blue in air).
- Derivation: in the Huygens construction, in the time the wavefront takes to move BC = v₁t in medium 1, it moves AE = v₂t in medium 2; the number of waves per second is the same, so λ₁/λ₂ = v₁/v₂ = n₂/n₁.
Conclusion: (a) reflected light: λ = 589 nm, ν = 5.09 × 10¹⁴ Hz, v = 3.00 × 10⁸ m s⁻¹; (b) refracted light: ν = 5.09 × 10¹⁴ Hz (unchanged), v = 2.26 × 10⁸ m s⁻¹, λ = 443 nm.
What is the shape of the wavefront in each of the following cases: (a) Light diverging from a point source. (b) Light emerging out of a convex lens when a point source is placed at its focus. (c) The portion of the wavefront of light from a distant star intercepted by the Earth.
Concept and context. A wavefront is the locus of points that oscillate in the same phase (a surface of constant phase); the energy of the wave travels along rays perpendicular to the wavefront. A point source in a uniform medium emits spherical wavefronts (NCERT Fig. 10.1a); far from the source a small portion of the sphere is practically a plane wavefront (Fig. 10.1b). Optical elements change the shape of the wavefront: a convex lens converts a plane wave into a converging spherical wave (and conversely a diverging spherical wave from the focus into a plane wave) because the central part of the wavefront, passing through the thickest glass, is delayed the most (Fig. 10.7b).
| Case | Shape of the wavefront | Reason |
|---|---|---|
| (a) Light diverging from a point source | spherical (concentric spheres centred on the source) | all points at the same distance from the source oscillate in phase; the radius increases with time at the speed of light |
| (b) Light emerging from a convex lens with the point source at its focus | plane | the lens converts the diverging spherical wavefront into a parallel beam; the rays are parallel, so the wavefronts perpendicular to them are planes |
| (c) The portion of the wavefront from a distant star intercepted by the Earth | plane | the star is at a very large distance, so the radius of the spherical wavefront is huge compared with the size of the Earth; a small portion of a very large sphere is practically a plane |
- Detail for (b): the source at the focus emits a spherical wave; the central part meets the thickest glass and is delayed most, the outer parts less, so the emergent wavefront is flat — rays emerge parallel to the axis (the reverse of the converging effect of Fig. 10.7b).
- Detail for (c): the radius of curvature of the wavefront equals the distance of the star (light-years); the Earth’s diameter (1.3 × 10⁷ m) is utterly negligible, so the wavefront is flat to an excellent approximation; this is why telescope mirrors are designed for parallel incident rays.
- Rays and wavefronts: rays are always perpendicular to wavefronts — radial for spherical and parallel for plane wavefronts.
Conclusion: (a) spherical; (b) plane; (c) plane.
(a) The refractive index of glass is 1.5. What is the speed of light in glass? (Speed of light in vacuum is 3.0 × 10⁸ m s⁻¹) (b) Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism?
Concept and context. The refractive index of a medium is the ratio of the speed of light in vacuum to its speed in the medium: n = c/v. A higher refractive index means that light travels more slowly in the medium. According to the wave theory, a wave bending towards the normal on entering a medium has a lower speed in it — as confirmed by Foucault’s experiment (1850) that light travels slower in water than in air. The refractive index of a transparent medium is not the same for all colours (wavelengths): it increases as the wavelength decreases — this variation is called dispersion and explains the splitting of white light into colours by a prism.
(a) Speed of light in glass
v = c/n = (3.0 × 10⁸ m s⁻¹)/1.5 = 2.0 × 10⁸ m s⁻¹
(b) Dependence on colour
- No — the speed of light in glass depends on the colour (frequency) of the light, because the refractive index of glass is different for different colours: n_violet > n_red (for typical crown glass n_red ≈ 1.51, n_violet ≈ 1.53).
- Hence violet light travels slower than red light in a glass prism (v = c/n is smaller for the larger n), so it is bent more (deviated more) by the prism: violet is deviated the most and red the least, which produces the spectrum. In vacuum all colours travel at the same speed c.
| Colour | Refractive index (typical crown glass) | Speed v = c/n |
|---|---|---|
| Red | ≈ 1.51 | ≈ 1.99 × 10⁸ m s⁻¹ |
| (mean) given | 1.50 | 2.0 × 10⁸ m s⁻¹ |
| Violet | ≈ 1.53 | ≈ 1.96 × 10⁸ m s⁻¹ |
- Cause (microscopic): light interacts with the atoms of the medium as forced oscillations; the response (and hence the effective speed) depends on the frequency of the light relative to the natural frequencies of the atoms (nearer to resonance for violet): so n increases with frequency in the visible range (normal dispersion).
- Consequences: chromatic aberration of lenses, dispersion of white light, rainbows.
Conclusion: (a) the speed of light in glass is 2.0 × 10⁸ m s⁻¹; (b) it is not independent of colour — violet light travels slower than red light in a glass prism.
In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
Concept and context. In Young’s double-slit experiment (NCERT Fig. 10.12) monochromatic light of wavelength λ passes through two narrow slits S₁ and S₂ separated by a distance d; being derived from the same source they are coherent. On a screen at distance D (D ≫ d) the waves interfere. For a point at distance x from the central point O the path difference between the two waves is Δ = S₂P − S₁P ≈ xd/D. Bright fringes (constructive interference) occur where Δ = nλ: x_n = nλD/d (n = 0, ±1, ±2, …); dark fringes where Δ = (n + ½)λ: x = (n + ½)λD/d. The fringes are equally spaced with fringe width β = λD/d.
Given: d = 0.28 mm = 0.28 × 10⁻³ m; D = 1.4 m; distance of the 4th bright fringe from the central bright fringe y₄ = 1.2 cm = 1.2 × 10⁻² m (n = 4).
- Position of the n-th bright fringe: y_n = nλD/d ⇒ λ = y_n d/(nD).
- Substitute: λ = (1.2 × 10⁻² × 0.28 × 10⁻³)/(4 × 1.4) = (3.36 × 10⁻⁶)/(5.6) m.
- λ = 6.0 × 10⁻⁷ m = 600 nm (orange-yellow light).
λ = y₄ d / (4D) = 600 nm
| Quantity | Value |
|---|---|
| d | 0.28 mm |
| D | 1.4 m |
| y₄ | 1.2 cm |
| Fringe width β = y₄/4 | 3.0 mm |
| λ = βd/D | 600 nm |
- Check via fringe width: β = y₄/4 = 3.0 mm; λ = β d/D = (3.0 × 10⁻³ × 0.28 × 10⁻³)/1.4 = 6.0 × 10⁻⁷ m ✓.
- Why “central bright” fringe to the fourth: the central fringe is the zero-order maximum (n = 0, path difference zero); the fourth bright fringe corresponds to a path difference of 4λ.
- The numbers show how Young could measure the tiny wavelength of light from large, measurable fringe separations (millimetres) using closely spaced slits.
Conclusion: the wavelength of the light used is 600 nm (6.0 × 10⁻⁷ m).
In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. What is the intensity of light at a point where path difference is λ/3?
Concept and context. In the double-slit experiment two coherent waves of equal amplitude a (individual intensity I₀ ∝ a²) arrive at a point P on the screen with a phase difference φ related to the path difference Δ by φ = (2π/λ)Δ. By the principle of superposition the resultant displacement is y = a cos ωt + a cos(ωt + φ) = 2a cos(φ/2) cos(ωt + φ/2), with amplitude 2a cos(φ/2). Since intensity ∝ (amplitude)², the intensity at P is I = 4I₀ cos²(φ/2) (NCERT Section 10.4, Fig. 10.10). The maximum intensity 4I₀ occurs when φ = 0, 2π, 4π … (Δ = 0, λ, 2λ …) and the minimum, zero, when φ = π, 3π … (Δ = λ/2, 3λ/2 …).
Given: at Δ₁ = λ the intensity is K; find the intensity at Δ₂ = λ/3.
- Phase difference at Δ₁ = λ: φ₁ = (2π/λ)λ = 2π ⇒ cos²(φ₁/2) = cos²π = 1 ⇒ I₁ = 4I₀ = K ⇒ I₀ = K/4 (intensity of each source separately).
- Phase difference at Δ₂ = λ/3: φ₂ = (2π/λ)(λ/3) = 2π/3 ⇒ φ₂/2 = π/3 ⇒ cos²(π/3) = (1/2)² = 1/4.
- Intensity at Δ₂: I₂ = 4I₀ cos²(φ₂/2) = 4I₀ × (1/4) = I₀ = K/4.
I = 4I₀ cos²(π/3) = I₀ = K/4
| Path difference Δ | Phase difference φ | cos²(φ/2) | Intensity |
|---|---|---|---|
| 0 | 0 | 1 | 4I₀ = K (central maximum) |
| λ/3 | 2π/3 | 1/4 | I₀ = K/4 |
| λ/2 | π | 0 | 0 (dark) |
| λ | 2π | 1 | 4I₀ = K (maximum) |
- Check: intensities at Δ = λ/3 and Δ = 2λ/3 are equal (both K/4); the average intensity over a fringe is 2I₀ = K/2 — the total energy is only redistributed (no gain or loss).
- Alternative: I = K cos²(πΔ/λ) with K the maximum intensity: at Δ = λ/3: K cos²(π/3) = K/4 ✓.
Conclusion: the intensity at the point where the path difference is λ/3 is K/4.
A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment. (a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm. (b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
Concept and context. In Young’s experiment the n-th bright fringe for wavelength λ lies at y_n = nλD/d, where d is the slit separation and D the distance to the screen; each wavelength forms its own fringe system with its own fringe width β = λD/d. For a beam containing two wavelengths, the bright fringes coincide where n₁λ₁ = n₂λ₂ for integers n₁, n₂ (same position, so the same path difference is an integral multiple of both wavelengths). The least such position corresponds to the smallest integers satisfying n₁/n₂ = λ₂/λ₁. Note on the data: the printed exercise does not state D and d; the standard version of this problem uses a slit separation d = 2 mm and screen distance D = 120 cm (D/d = 600), which are used for the numerical results below — the general results are given in terms of D/d as well.
Given: λ₁ = 650 nm = 6.5 × 10⁻⁷ m; λ₂ = 520 nm = 5.2 × 10⁻⁷ m; (standard data) d = 2 mm = 2 × 10⁻³ m, D = 120 cm = 1.2 m ⇒ D/d = 600.
(a) Distance of the third bright fringe for 650 nm
- y₃ = 3λ₁D/d = 3 × 6.5 × 10⁻⁷ × (1.2/2 × 10⁻³) = 3 × 6.5 × 10⁻⁷ × 600.
- y₃ = 1.17 × 10⁻³ m = 1.17 mm (general: y₃ = 1.95 × 10⁻⁶ × D/d metres).
(b) Least distance where bright fringes of the two wavelengths coincide
- Coincidence condition: n₁λ₁ = n₂λ₂ ⇒ n₁/n₂ = λ₂/λ₁ = 520/650 = 4/5.
- The smallest integers: n₁ = 4 (650 nm), n₂ = 5 (520 nm) — check: 4 × 650 = 5 × 520 = 2600 nm.
- Distance: y = n₁λ₁D/d = 4 × 6.5 × 10⁻⁷ × 600 = 1.56 × 10⁻³ m = 1.56 mm (general: y = 2.6 × 10⁻⁶ D/d metres). Beyond this the coincidences repeat at multiples (8th and 10th orders, at 3.12 mm, etc.).
(a) 3λ₁D/d = 1.17 mm ; (b) 4λ₁D/d = 5λ₂D/d = 1.56 mm (for D/d = 600)
| Order n | 650 nm: y_n (mm) | 520 nm: y_n (mm) |
|---|---|---|
| 1 | 0.39 | 0.312 |
| 2 | 0.78 | 0.624 |
| 3 | 1.17 | 0.936 |
| 4 | 1.56 | 1.248 |
| 5 | 1.95 | 1.56 |
- Why the first coincidence is at (4, 5): 650 and 520 have the ratio 5:4 (their highest common factor is 130 nm: 650 = 5 × 130 and 520 = 4 × 130); the path difference of 2600 nm = 20 × 130 nm is the least common multiple of the two wavelengths.
- Use of white light: only the central fringe is white and the fringes of different colours overlap and become coloured as the order increases.
Conclusion: (a) the third bright fringe of 650 nm lies 1.17 mm from the central maximum (3λ₁D/d, for D/d = 600); (b) the least distance at which the bright fringes of both wavelengths coincide is 1.56 mm (the 4th fringe of 650 nm coincides with the 5th of 520 nm).