Part 4 · Chapter 35

Interference

In the last chapter light traveled in tidy straight-line rays. Now we abandon that simplicity and return to light as a wave — because only waves can interfere. When two coherent light waves overlap, they can add to a brilliant maximum or cancel to darkness, depending entirely on the phase between them. This is the physics behind the shifting blue of a butterfly's wing, the colors of an oil slick, the rainbow's faint companion arcs, and Thomas Young's 1801 experiment that finally proved light is a wave. We build the toolkit step by step: how phase changes with path length, with the medium, and with reflection, and how those changes paint patterns of light and shadow we can actually measure.

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 80 min
i What you'll learn
  • Light as a wave via Huygens' principle, which rebuilds the laws of reflection and refraction and gives \(n = c/v\) its meaning.
  • The wavelength in a medium shrinks, \(\lambda_n = \dfrac{\lambda}{n}\), while the frequency stays the same; a path of length \(L\) in two media shifts the phase by \(N_2 - N_1 = \dfrac{L}{\lambda}(n_2 - n_1)\) wavelengths.
  • Young's double-slit experiment: the path-length difference \(\Delta L = d\sin\theta\) sets bright fringes at \(d\sin\theta = m\lambda\) and dark fringes at \(d\sin\theta = (m+\tfrac12)\lambda\), with even fringe spacing \(\Delta y = \dfrac{\lambda D}{d}\).
  • Coherence is the steady phase relation that makes a fringe pattern possible, and phasors give the intensity \(I = 4I_0\cos^{2}\tfrac12\phi\) with \(\phi = \dfrac{2\pi d}{\lambda}\sin\theta\).
  • Thin-film interference: a reflection off a higher index adds half a wavelength ("higher means half"), giving for a film in air a bright film at \(2L = (m+\tfrac12)\dfrac{\lambda}{n_2}\) and a dark film at \(2L = m\dfrac{\lambda}{n_2}\) — the basis of anti-reflection coatings.
  • The Michelson interferometer turns a fringe count into a length measurement: \(N_m - N_a = \dfrac{2L}{\lambda}(n-1)\).
Section 35-1

What Is Physics?

One of the central goals of physics is to understand the nature of light — a goal made difficult, and rich, by how complicated light is. Nature has exploited optical interference for color long before we understood it: the wings of a Morpho butterfly are a dull brown, yet their top surface flashes an arresting blue produced not by pigment but by light waves interfering as they reflect, and the tint shifts as your viewing angle changes. The same color-shifting trick is printed into modern currency to foil copiers, which can match a color from one angle but not its shift. To understand any of this we must set aside the rays of geometrical optics and return to the wave nature of light.

Section 35-2

Light as a Wave

The first convincing wave theory of light came from Christiaan Huygens in 1678. Simpler than Maxwell's later electromagnetic theory but still useful today, it rests on a geometric rule for advancing a wavefront in time.

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Huygens' principle
Every point on a wavefront acts as a source of spherical secondary wavelets; a moment \(\Delta t\) later the new wavefront is the surface tangent to all those wavelets.

For a plane wave in vacuum the wavelets each grow by \(c\,\Delta t\), and their common tangent is a new plane parallel to the first — the wave simply advances. The power of the picture appears at a boundary: when one part of the wavefront enters glass and slows, its wavelets grow more slowly, the tangent plane tilts, and the wave bends. That single idea reproduces Snell's law.

Huygens' construction gives the law of refraction
\[ \frac{\lambda_1}{\lambda_2} = \frac{v_1}{v_2}, \qquad n = \frac{c}{v}, \qquad n_1\sin\theta_1 = n_2\sin\theta_2 \]
Wavelengths in two media are proportional to the wave speeds there. Defining the index of refraction \(n = c/v\) and combining the two relations reproduces Snell's law from Chapter 33 — now derived, not assumed, from waves.
Section 35-3

Wavelength and Phase in a Medium

Because speed changes in a medium, so does wavelength — and that is the seed of every interference effect that follows. If light of vacuum wavelength \(\lambda\) enters a medium of index \(n\), its wavelength shrinks, yet its frequency is unchanged: the medium cannot alter how many wave crests per second pass a point, only how tightly they are packed.

Wavelength in a medium (frequency unchanged)
\[ \lambda_n = \frac{\lambda}{n}, \qquad f_n = f \]
The larger the index, the shorter the wavelength inside. The frequency is fixed by the source, so only the wavelength and speed differ from their vacuum values.

Now send two initially in-phase waves through equal lengths \(L\) of two different media. Because their wavelengths differ inside, they fit a different number of cycles into the same distance and emerge with a new phase difference — counted in wavelengths, it is the difference of those two cycle counts.

Phase difference from an index difference
\[ N_2 - N_1 = \frac{L}{\lambda}\,(n_2 - n_1) \]
This counts the extra wavelengths gained along the higher-index path. Only the fractional part matters for interference: a whole-number shift returns the waves to phase, an extra \(0.5\) wavelength makes them fully out of phase (darkness), and \(0\) or \(1.0\) makes them fully in phase (brightness). Recall \(1\) wavelength \(= 2\pi\,\text{rad} = 360^\circ\).
Rainbows are interference too. Light enters a raindrop all along the side facing the Sun, and different parts travel different paths inside, emerging with different phases. At certain angles the emerging waves of a given color are in phase and interfere constructively — that is the rainbow. The faint extra arcs sometimes seen just inside a primary bow, the supernumeraries, are further constructive-interference maxima, a naturally occurring proof that light is a wave.
Section 35-4

Diffraction

Before Young's experiment we need one more wave idea. When a wave meets a barrier with an opening about the size of its wavelength, the part passing through flares out into the region beyond — it diffracts — exactly as Huygens' wavelets would predict. The narrower the slit, the more pronounced the spreading. Diffraction happens for waves of every kind; it is also why geometrical optics fails for apertures comparable to the wavelength, since any attempt to pin light into a perfect ray with a tiny slit only makes it spread more.

Section 35-5

Young's Interference Experiment

In 1801 Thomas Young proved light is a wave by showing it interferes. Monochromatic light first passes through a single slit \(S_0\), which by diffraction becomes a coherent point source; its light then spreads to two slits \(S_1\) and \(S_2\), and the two diffracted waves overlap beyond the screen, producing alternating bright and dark bands — interference fringes — on a viewing screen. Young even measured the average wavelength of sunlight, getting 570 nm against the modern 555 nm.

The waves leave the two slits in phase but travel different distances to a point \(P\) at angle \(\theta\). With the screen far away (\(D \gg d\)) the two rays are essentially parallel, and the geometry gives a simple path-length difference.

Path-length difference
\[ \Delta L = d\sin\theta \]
Here \(d\) is the slit separation and \(\theta\) the angle to point \(P\) from the central axis. Everything about the pattern follows from comparing this \(\Delta L\) to the wavelength.
Maxima (bright) and minima (dark)
\[ d\sin\theta = m\lambda \quad(\text{maxima}), \qquad d\sin\theta = \left(m+\tfrac12\right)\lambda \quad(\text{minima}), \qquad m = 0,1,2,\dots \]
A whole number of wavelengths of path difference means the waves arrive in phase → a bright fringe; an odd number of half-wavelengths means out of phase → a dark fringe. The integer \(m\) is the order; \(m=0\) is the central bright fringe on the axis.
Fringe position and even spacing (small angles)
\[ y_m = \frac{m\lambda D}{d}, \qquad \Delta y = \frac{\lambda D}{d} \]
Using \(\sin\theta \approx \tan\theta = y/D\), the bright fringes sit at \(y_m\), and adjacent maxima are an equal distance \(\Delta y\) apart — the fringes are evenly spaced near the center, independent of \(m\).
Section 35-6

Coherence

A fringe pattern appears only if the phase difference between the two arriving waves is steady in time — the light must be coherent. In Young's setup the two slits are fed by one wave, so their phase difference is fixed and the light is fully coherent. Replace the slits with two independent bulbs and the story collapses: the countless atoms in each filament emit randomly for only nanoseconds, so the phase difference jitters far faster than any eye or detector can follow, and the interference washes out to uniform illumination. Sunlight is only partially coherent (the faint speckle on a fingernail in bright sun is interference from nearby points). A laser is the opposite extreme — its atoms emit cooperatively, giving light that is coherent, nearly monochromatic, and tightly beamed.

Section 35-7

Intensity in Double-Slit Interference

The fringe equations locate the bright and dark bands, but how bright is the screen at an arbitrary angle? Treat the two arriving fields as having equal amplitude \(E_0\) and a fixed phase difference \(\phi\), then add them. The cleanest route is the phasor method: represent each wave by a rotating vector of length \(E_0\) and add the vectors tip-to-tail. The resultant amplitude is \(E = 2E_0\cos\tfrac12\phi\), and intensity goes as amplitude squared.

Double-slit intensity
\[ I = 4I_0\cos^{2}\tfrac12\phi, \qquad \phi = \frac{2\pi d}{\lambda}\sin\theta \]
Here \(I_0\) is the intensity from one slit alone, and \(\phi\) is the phase difference, just \(2\pi/\lambda\) times the path difference \(d\sin\theta\). Setting the cosine to its extremes recovers the maxima (\(d\sin\theta = m\lambda\)) and minima (\(d\sin\theta = (m+\tfrac12)\lambda\)) found earlier.
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Interference redistributes energy — it never creates or destroys it
The intensity swings from \(0\) at the dark fringes to \(4I_0\) at the bright ones, yet the average across the screen stays \(2I_0\) — exactly the two-slit total.

Coherent light merely shuffles the energy into bright and dark bands; averaged over the screen it equals the \(2I_0\) you would get from two incoherent sources. The phasor toolkit also handles three or more waves: draw the phasors head to tail, and the closing vector gives the resultant amplitude and phase.

Section 35-8

Interference from Thin Films

The colors of a soap bubble or an oil slick come from light reflecting off the front and back surfaces of a film only a wavelength or so thick, then interfering. Three things can change the phase difference between those two reflected waves: the path length (the back-surface wave crosses the film twice, an extra \(2L\)), the wavelength inside the film (\(\lambda_{n_2} = \lambda/n_2\)), and — crucially — reflection itself.

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Reflection phase shifts — "higher means half"
Light reflecting off a medium of higher index flips by half a wavelength; reflecting off a lower index, it is unshifted. Refraction never shifts the phase.

Like a wave pulse inverting when it hits a heavier string, a light wave inverts (a \(\pi\)-rad, half-wavelength shift) when it reflects from the higher-index side, and reflects unchanged from the lower-index side. For a film in air, the front reflection gains \(0.5\) wavelength and the back reflection gains none — so the two reflected waves start out already half a wavelength apart before the path difference is even counted.

Thin film in air (light reflected): bright and dark
\[ 2L = \left(m+\tfrac12\right)\frac{\lambda}{n_2} \quad(\text{bright}), \qquad 2L = m\,\frac{\lambda}{n_2} \quad(\text{dark}), \qquad m = 0,1,2,\dots \]
Valid for near-normal incidence with the same medium (air) on both sides, so the single \(0.5\)-wavelength reflection shift applies. Watch the surroundings: if both reflections shift by the same amount (or neither does), the two equations swap roles — always build the reflection-shift table first.
Two everyday consequences. A vertical soap film drains thicker at the bottom; its very top can become so thin (\(L \ll \lambda\)) that only the reflection shift matters, leaving the two waves half a wavelength apart and the top dark, even in bright light. And an anti-reflection coating — a film like MgF₂ on a lens — is chosen so the two reflected waves are exactly out of phase and cancel, killing the reflection at a chosen wavelength so more light passes through.
Section 35-9

Michelson's Interferometer

An interferometer measures lengths to a fraction of a wavelength by counting fringes. In Michelson's 1881 design, a beam splitter divides a wave into two; each half travels down an arm, reflects off a mirror, and returns to recombine at the telescope. The path-length difference is \(2d_2 - 2d_1\), so moving one mirror by half a wavelength shifts the pattern by one whole fringe.

Fringe shift from inserting a thin material
\[ N_m - N_a = \frac{2L}{\lambda}\,(n - 1) \]
Placing a slab of thickness \(L\) and index \(n\) in one arm changes the number of wavelengths along the round trip by this amount; counting the fringes that sweep past then yields \(L\) in terms of \(\lambda\). With this instrument Michelson expressed the standard meter in wavelengths of light (1907 Nobel Prize), foreshadowing today's definition of the meter through the speed of light.
Worked Examples

Putting It to Work

1 Spacing of double-slit fringes

Problem. Light of wavelength \(\lambda = 546\,\mathrm{nm}\) falls on two slits separated by \(d = 0.12\,\mathrm{mm}\). The screen is \(D = 55\,\mathrm{cm}\) away. What is the distance between adjacent bright fringes near the center?

Solution. Near the center the angles are small, so the fringes are evenly spaced by \(\Delta y = \lambda D/d\).

Δy = λD/d
\[ \Delta y = \frac{(546\times10^{-9})(0.55)}{0.12\times10^{-3}} \approx 2.5\times10^{-3}\,\mathrm{m} = 2.5\,\mathrm{mm} \]

The spacing grows with wavelength and screen distance and shrinks as the slits move apart — push the slits closer and the pattern spreads out, the lever that makes the tiny wavelength of light visible to the naked eye.

2 Phase shift from a plastic layer

Problem. Two in-phase waves of wavelength \(\lambda = 550.0\,\mathrm{nm}\) travel equal lengths; one through air (\(n_1 = 1.000\)), the other through a plastic strip of index \(n_2 = 1.600\) and thickness \(L = 2.600\,\mu\mathrm{m}\). Find their phase difference in wavelengths and its effective value.

Solution. Count the extra wavelengths along the plastic path with \(N_2 - N_1 = \dfrac{L}{\lambda}(n_2 - n_1)\).

N₂ − N₁ = (L/λ)(n₂ − n₁)
\[ N_2 - N_1 = \frac{2.600\times10^{-6}}{5.500\times10^{-7}}\,(1.600 - 1.000) \approx 2.84 \]

So the waves differ by 2.84 wavelengths. Only the fractional part governs the interference, giving an effective phase difference of 0.84 wavelength — between \(0.5\) (fully dark) and \(1.0\) (fully bright), but close to \(1.0\), so the waves would meet in nearly constructive interference and produce a fairly bright spot. (Note: take the decimal of the answer in wavelengths, never of the radian value.)

3 Brightest color from a water film

Problem. White light (400–690 nm) strikes a water film (\(n_2 = 1.33\)) of thickness \(L = 320\,\mathrm{nm}\) suspended in air, at near-normal incidence. At what visible wavelength is the reflected light brightest?

Solution. Air on both sides means one \(0.5\)-wavelength reflection shift, so the bright-film condition is \(2L = (m+\tfrac12)\lambda/n_2\). Solve for \(\lambda\).

λ = 2n₂L / (m + ½)
\[ \lambda = \frac{2n_2 L}{m+\tfrac12} = \frac{2(1.33)(320\,\mathrm{nm})}{m+\tfrac12} = \frac{851\,\mathrm{nm}}{m+\tfrac12} \]

For \(m=0\), \(\lambda = 1700\,\mathrm{nm}\) (infrared); for \(m=1\), \(\lambda \approx 567\,\mathrm{nm}\) (yellow-green, visible); for \(m=2\), \(\lambda \approx 340\,\mathrm{nm}\) (ultraviolet). Only \(567\,\mathrm{nm}\) lands in the visible band, so the film looks brightest in yellow-green.

4 Anti-reflection coating

Problem. A glass lens (\(n_3 = 1.50\)) is coated with magnesium fluoride (\(n_2 = 1.38\)) to cancel reflection at \(\lambda = 550\,\mathrm{nm}\). Find the least coating thickness.

Solution. At the air–MgF₂ surface the reflection shifts by \(0.5\) wavelength; at the MgF₂–glass surface it also shifts by \(0.5\) (since \(1.50 > 1.38\)). Equal shifts cancel, so to make the two reflected waves cancel we need the path term to supply an odd number of half-wavelengths: \(2L = (m+\tfrac12)\lambda/n_2\).

L = λ / 4n₂ (least, m = 0)
\[ L = \frac{\lambda}{4 n_2} = \frac{550\,\mathrm{nm}}{4(1.38)} \approx 99.6\,\mathrm{nm} \]

A quarter-wave layer (measured inside the coating) does the job. Because both surfaces shift equally here, the algebra flips relative to the soap film of Example 3 — a reminder to always tabulate the reflection shifts before choosing the bright or dark equation.

Review

Chapter Summary

✓ Huygens & refraction

Wavefronts advance as tangents to secondary wavelets; this gives \(n = c/v\) and \(n_1\sin\theta_1 = n_2\sin\theta_2\).

✓ Wavelength & phase

\(\lambda_n = \dfrac{\lambda}{n}\), \(f_n = f\); index difference shifts phase by \(\dfrac{L}{\lambda}(n_2-n_1)\) wavelengths.

✓ Path difference

\(\Delta L = d\sin\theta\) sets everything; only its fractional part (in \(\lambda\)) matters.

✓ Fringes

Bright \(d\sin\theta = m\lambda\); dark \(d\sin\theta = (m+\tfrac12)\lambda\); spacing \(\Delta y = \dfrac{\lambda D}{d}\).

✓ Coherence

A steady phase relation is required for fringes; lasers are coherent, independent bulbs are not.

✓ Intensity

\(I = 4I_0\cos^{2}\tfrac12\phi\), \(\phi = \dfrac{2\pi d}{\lambda}\sin\theta\); average \(2I_0\), max \(4I_0\).

✓ Thin films

"Higher means half." Film in air: bright \(2L = (m+\tfrac12)\dfrac{\lambda}{n_2}\), dark \(2L = m\dfrac{\lambda}{n_2}\).

✓ Michelson

Counting fringes measures length: \(N_m - N_a = \dfrac{2L}{\lambda}(n-1)\).

Practice

Problems

Unless stated otherwise, take air to have \(n = 1.00\) and assume near-normal incidence for thin films. For interference of any two waves, work in wavelengths and keep only the fractional part: \(0\) or \(1.0\) is fully constructive, \(0.5\) fully destructive. For thin films, build the reflection-shift table ("higher means half") before choosing the bright or dark equation.

  1. The wavelength of yellow sodium light in air is 589 nm. (a) What is its frequency? (b) What is its wavelength in glass of index 1.52? (c) From these, find its speed in the glass.
  2. The speed of light in a certain liquid is \(1.92\times10^{8}\,\mathrm{m/s}\). What is the index of refraction of the liquid?
  3. Two waves of light in air, of wavelength 400 nm, are initially in phase. One passes through a glass layer (\(n_1 = 1.60\)) of thickness \(L\); the other through an equally thick plastic layer (\(n_2 = 1.50\)). What is the smallest \(L\) that leaves them with a phase difference of 5.65 rad?
  4. In a double-slit arrangement the slits are separated by 100 times the wavelength of the light. (a) What is the angular separation between the central maximum and an adjacent maximum? (b) What is their separation on a screen 50.0 cm away?
  5. Young's experiment is performed with blue-green light of wavelength 500 nm. The slits are 1.20 mm apart and the screen is 5.40 m away. How far apart are the bright fringes near the center?
  6. Monochromatic green light (550 nm) illuminates two slits 7.70 μm apart. Find the angular deviation of the third-order (\(m=3\)) bright fringe (a) in radians and (b) in degrees.
  7. A double-slit arrangement produces fringes for sodium light (589 nm) that are \(0.20^\circ\) apart. What is the angular fringe separation if the whole apparatus is immersed in water (\(n = 1.33\))?
  8. In a double-slit experiment \(d = 5.0\,\mathrm{mm}\) and the screen is 1.0 m away. Two patterns appear, one from 480 nm light and one from 600 nm. What is the separation on the screen between their third-order (\(m=3\)) bright fringes?
  9. Two waves of the same frequency have amplitudes 1.00 and 2.00 and interfere where their phase difference is \(60.0^\circ\). Find the resultant amplitude. (Hint: use phasors.)
  10. Add by the phasor method: \(y_1 = 10\sin\omega t\), \(y_2 = 15\sin(\omega t + 30^\circ)\), \(y_3 = 5.0\sin(\omega t - 45^\circ)\).
  11. A thin flake of mica (\(n = 1.58\)) covers one slit of a double slit. The central point on the screen is now occupied by what had been the seventh bright fringe (\(m=7\)). If \(\lambda = 550\,\mathrm{nm}\), find the mica's thickness.
  12. We wish to coat flat glass (\(n = 1.50\)) with a transparent film (\(n = 1.25\)) so that reflection at 600 nm is eliminated by interference. What is the minimum coating thickness?
  13. Light of wavelength 624 nm is incident perpendicularly on a soap film (\(n = 1.33\)) suspended in air. Find the (a) least and (b) second-least thickness for which the reflection is fully constructive.
  14. The rhinestones in costume jewelry are glass (\(n = 1.50\)) coated with silicon monoxide (\(n = 2.00\)) to make them more reflective. Find the minimum coating thickness for fully constructive reflection of 560 nm light at normal incidence. (Careful with the reflection shifts.)
  15. Moving mirror \(M_2\) of a Michelson interferometer through 0.233 mm produces a shift of 792 bright fringes. What is the wavelength of the light?
  16. A thin film (\(n = 1.40\)) placed in one arm of a Michelson interferometer, perpendicular to the path, shifts the pattern by 7.0 bright fringes for 589 nm light. Find the film thickness.
Tip: three habits carry this whole chapter. First, do every phase bookkeeping in wavelengths and keep only the fractional part — \(0\) or \(1.0\) is a bright maximum, \(0.5\) a dark minimum, and you convert to radians (\(\times 2\pi\)) only at the very end, never taking the decimal part of a radian value. Second, for two-slit problems the single quantity that decides everything is the path-length difference \(\Delta L = d\sin\theta\), so find it first, compare it to \(\lambda\), and let \(d\sin\theta = m\lambda\) or \((m+\tfrac12)\lambda\) tell you bright or dark — and remember the wavelength shrinks to \(\lambda/n\) if the apparatus sits in water or oil. Third, for thin films never reach for a formula until you have written down the reflection phase shift at each surface using "higher means half": when the two shifts differ by half a wavelength the bright condition is \(2L = (m+\tfrac12)\lambda/n_2\), but when they are equal (or both absent) the bright and dark equations trade places.