Wave Optics — Interference, Diffraction & Thin Films

For interference patterns to be stable and visible, the two sources must be coherent — they must maintain a constant phase relationship. Two separate light bulbs are incoherent: their phases randomly drift, so any interference pattern averages out to uniform brightness in milliseconds.

Coherent sources are created by splitting a single source into two (Young's double slit, thin films). Lasers are inherently coherent. Ordinary light from slits works because the slit selects a narrow region of the wavefront that IS coherent with itself over the distance between slits.

This is why interference from two separate lasers of different frequencies gives no stable fringes — but two slits illuminated by one laser give beautiful, stable fringes.

The sum x + y tells you the total distance both waves travel, but that's irrelevant. What matters is whether the two waves arrive at point P in phase or out of phase. A wave from source A that traveled 10λ has oscillated exactly 10 full cycles and arrives in the same phase as when it left A (since it started in phase with B, it's still in phase with whatever B's wave is doing). The only thing that shifts their relative phase is the extra distance one traveled over the other — the difference.

Analogy: two runners start a race at the same time from the same place. One runs 10 laps, the other runs 9 laps, both in the same time. They meet at the finish when their lap counts differ by an integer — not when their total laps sum to some value.

A sinusoidal wave completes one full oscillation per wavelength λ. If wave 2 travels exactly λ/2 farther than wave 1 before reaching point P, it is exactly half a cycle behind. Wave 1 is at its peak (+A) exactly when wave 2 is at its trough (−A). They superpose to give A + (−A) = 0. Perfect cancellation.

At δ = 3λ/2, wave 2 is 1½ cycles behind — which is the same as ½ cycle out of phase (one full cycle of extra travel brings it back in sync, then the remaining ½ cycle flips it). So destructive happens at ALL half-odd multiples: λ/2, 3λ/2, 5λ/2, ...

If a problem gives you the screen distance L, slit separation d, and wavelength λ, you can find everything: fringe position (y = mλL/d), fringe spacing (Δy = λL/d), angle to any fringe (sinθ = mλ/d), and which m values are bright or dark at a given position.

Common exam move: "Find the separation d given that the 3rd bright fringe is at y = 2.4 cm, L = 1.5 m, λ = 600 nm." Use y = mλL/d → d = mλL/y = 3 × 600e-9 × 1.5 / 0.024 = 1.125 × 10⁻⁴ m = 0.1125 mm.

Imagine the slit of width a divided into many thin strips. For each strip in the top half of the slit, there's a corresponding strip in the bottom half, exactly a/2 farther from the slit edge. When d sinθ = λ/2 for pairs a/2 apart — which happens when a sinθ = λ — every strip in the top half is paired with a strip in the bottom half that is exactly λ/2 out of phase. Every pair cancels. The whole slit contributes zero amplitude → dark fringe at m=1.

For m=2, divide the slit into four quarters. Each quarter cancels the quarter below it. This gives a dark fringe at a sinθ = 2λ. And so on for higher m.

The central maximum (m=0) is special: all wavelets arrive roughly in phase because the path differences across the slit are much less than λ/2, so they add constructively rather than canceling.

Verify: a = λL/y₁ = (600 × 10⁻⁹ m × 3 m) / (2.25 × 10⁻³ m) = 1.8 × 10⁻⁶ / 2.25 × 10⁻³ = 8 × 10⁻⁴ m = 0.8 mm. This is physically reasonable: a slit of 0.8 mm illuminated by 600 nm visible light would produce a central maximum about 4.5 mm wide on a screen 3 m away — completely consistent with a typical lab setup.

The small angle check: θ₁ = arctan(2.25/3000) = 0.043° — well within the small angle regime (sinθ ≈ tanθ is valid for angles below ~10°). The approximation is excellent here.

Diffraction happens because waves bend around edges. The degree of bending depends on how large the wavelength is compared to the slit width. A longer wavelength diffracts more (spreads out more); a shorter wavelength diffracts less (stays more concentrated near the center).

In water, the wavelength of light is shorter — each wave cycle packs into less space. The light "sees" the slit as relatively wider (more slit widths per wavelength). So it diffracts less. The pattern compresses. This is true for ALL diffraction and interference phenomena: submerge in denser medium → smaller λ → pattern shrinks by factor n.

What does NOT change: the frequency of the light (same color to your eyes at the screen edge), the slit width a, and the screen distance L. Only λ changes.

This comes from the electromagnetic wave boundary conditions. When an EM wave hits a boundary with a higher index of refraction (= denser optical medium = slower light), the reflected wave undergoes a 180° phase shift. This is mathematically equivalent to adding an extra path of λ/2.

Analogy with mechanical waves: attach a string to a heavy rope. When a pulse on the string reaches the heavy rope, the reflected pulse is inverted (flipped upside down). The heavy rope is "optically denser." If instead the rope is attached to a lighter string, the reflected pulse is not inverted — same as going from high-n to low-n in optics with no phase shift.

Key consequence: if BOTH reflections have phase shifts, they cancel (net zero phase shift from reflections). If ONLY ONE reflection has a phase shift, it contributes an extra λ/2 to the effective path difference. You must always count the phase shifts carefully before deciding if the total path difference is constructive or destructive.

Step 1: Identify the three media (top, film, bottom) and their indices n₁ (above film), n₂ (film), n₃ (below film).

Step 2: Ray 1 reflects at the top boundary. Does n increase going in (n₁ → n₂, n₂ > n₁)? If yes: 180° shift. If no: no shift.

Step 3: Ray 2 reflects at the bottom boundary. Does n increase going in (n₂ → n₃, n₃ > n₂)? If yes: 180° shift. If no: no shift.

Step 4: Count net shifts. 0 or 2 shifts → conditions as for zero phase difference. 1 shift → conditions swap (constructive uses the half-integer formula).

Step 5: Use λ_film = λ₀/n₂ in the path difference equation.

Ray 1 reflects from air→benzene and gets a 180° flip (equivalent to adding λ/2 to its phase). Ray 2 has no reflection flip at the bottom. For them to arrive back at the top surface in phase (constructive), Ray 2's extra path (= 2t in the film, measured in wavelengths of the film = 2t·n/λ₀) must COMPENSATE for the λ/2 phase difference caused by Ray 1's flip.

So: 2t·n/λ₀ = ½ + m (for constructive). This gives 2t = (m + ½)λ₀/n. For minimum thickness, m=0: 2t = λ₀/(2n), so t = λ₀/(4n) = 575/(4×1.50) nm = 95.8 nm.

This formula t = λ/(4n) for minimum thickness of a film with one phase shift is worth memorizing — it appears in anti-reflection coatings, quarter-wave plates, and many optics applications.

A soap bubble has varying thickness from top (gravity drains it thin) to sides (thicker). Different thicknesses give constructive reflection for different wavelengths of white light. The top appears black (zero thickness → path difference near 0 → destructive for all λ because of the one-phase-shift geometry). Partway down: constructive for blue. Lower: constructive for green, then yellow, then red. This is the rainbow of colors you see on a soap bubble — all from a film only a few hundred nanometers thick.

A CD is the same principle but with microscopic pits and lands: the reflections from pit edges and flat surfaces interfere, and the spacing is engineered so you get a rainbow of colors at different angles.

An anti-reflection coating is optimized for one wavelength (usually 550 nm = green). At 550 nm the destructive reflection is nearly complete. But at shorter wavelengths (blue, ~450 nm) and longer wavelengths (red, ~700 nm), the coating is the wrong thickness and doesn't cancel the reflection perfectly. So some blue and red light still reflects — and blue + red = purple/magenta. This is why camera lenses, microscope objectives, and quality eyeglasses appear to have a purple or green sheen when you look at them in white light.

In a medium, light travels slower: v = c/n. The frequency f is set by the source and doesn't change when light enters a medium (the EM oscillation frequency must match at the boundary — otherwise charge builds up). Since v = fλ, and f is constant but v decreases, λ must decrease proportionally: λ_medium = v/f = (c/n)/f = λ₀/n.

This means light of 500 nm green in air is about 376 nm "blue" wavelength inside glass with n=1.33. But your eye doesn't detect the wavelength inside the glass — it detects the frequency, which determines color. So green light is still green to your brain even though the wavelength inside the glass is shorter.

For diffraction and interference, what matters is the wavelength in the medium where the waves travel — the physical spacing between wave crests in space. Using the shorter λ in all formulas gives the correct fringe positions.

In geometric optics, a slit casts a sharp shadow with width exactly equal to a. But in wave optics, diffraction spreads the light beyond the slit's geometric shadow. The central maximum has width 2λL/a, which can be much larger than a itself if a is small (comparable to λ).

When a >> λ (much wider than the wavelength), 2λL/a ≈ 0 — the central maximum becomes very narrow and we recover the geometric shadow. This is why you don't see diffraction from everyday slits: typical slits are millimeters wide, and visible light is 500 nm, so 2λL/a is extremely small for normal distances L.

But a slit of 0.5 mm width illuminated by 500 nm light with L=1 m gives a central max width of 2×500e-9×1/5e-4 = 2 mm — visible and measurable. This is the regime of the textbook experiments and exam problems.

Every formula in this chapter answers the same question: when two waves from different paths meet at the same point, do they reinforce or cancel? The answer depends entirely on the total phase difference, which comes from two sources: (1) the path length difference δ_path = 2t or d sinθ, expressed as a fraction of the wavelength; and (2) any 180° flips from reflections off denser media.

If the total phase is a whole number of λ → constructive. Half-odd number of λ → destructive. This is Huygens' principle applied systematically. Huygens said every point on a wavefront is a source of secondary wavelets. Where those wavelets constructively interfere, you see light. Where they cancel, you see darkness.

The big picture: the double slit (Young, 1801) was the first proof that light is a wave. Before Young's experiment, Newton's corpuscular model of light (particles) was dominant. Young showed that light produces interference fringes — something only waves do. This experiment permanently settled the wave nature of light, 200 years before quantum mechanics showed it was also a particle. The exam question you're answering is a descendant of one of the most decisive experiments in the history of physics.