Difficulty: Intermediate | Prerequisites: Basic wave properties (wavelength, frequency, speed), trigonometry, understanding of superposition.
Wave optics is where light stops behaving like a straight-line ray and starts behaving like a wave. This unit covers what happens when light passes through narrow slits, bounces off thin films, or diffracts through a grating. If geometric optics (mirrors, lenses, Snell's law) tells you where light goes, wave optics tells you what it does when it gets there, and why you sometimes see bright and dark fringes instead of a uniform spot. You should already be comfortable with the idea that waves can add constructively or destructively, and with basic trig for resolving angles.
When light hits a diffraction grating, it splits into bright fringes at angles set by d sin θ = mλ. When light reflects off the top and bottom of a thin film, the two reflected beams interfere, and whether you get constructive or destructive interference depends on the film thickness, the wavelength inside the film, and whether either reflection picks up a phase shift. Single-slit diffraction produces a wide central bright fringe with progressively dimmer fringes on either side, with dark fringes at a sin θ = mλ.
Diffraction grating
An optical element with many equally spaced slits (or lines) that splits incoming light into multiple beams at specific angles. Think of it as a precision tool for separating wavelengths, like a much sharper version of a prism.
Order number (m)
The integer label for each bright fringe produced by a grating or double slit. m = 0 is the central maximum, m = 1 is the first-order fringe, m = 2 the second-order, and so on. In simple terms, it counts how many extra wavelengths of path difference exist between adjacent slits.
Grating spacing (d)
The distance between adjacent slits on a diffraction grating. If a grating has N lines per millimetre, d = 1/N. This is the number you plug into d sin θ = mλ.
Thin-film interference
The phenomenon where light reflected from the top and bottom surfaces of a thin transparent layer interferes with itself, producing bright or dark regions depending on the film thickness and the wavelength. Think of it as the reason you see rainbow colours in soap bubbles and oil slicks.
Phase shift on reflection
When light reflects off a surface where the refractive index increases (going from low n to high n), the reflected wave picks up an extra half-wavelength (π) phase shift. No extra phase shift occurs when reflecting off a surface where n decreases. In simple terms, bouncing off a denser medium flips the wave.
Constructive interference
Occurs when two waves arrive in phase (crest meets crest), reinforcing each other. The net path difference plus any phase shifts equals a whole number of wavelengths.
Destructive interference
Occurs when two waves arrive out of phase (crest meets trough), cancelling each other. The net path difference plus any phase shifts equals a half-integer number of wavelengths.
Anti-reflective coating
A thin film applied to a lens surface, designed so that the two reflected beams (from the top and bottom of the film) destructively interfere, reducing glare. Think of it as using interference to cancel out unwanted reflections.
Single-slit diffraction
The spreading of light after it passes through a single narrow opening, producing a broad central bright fringe flanked by progressively dimmer secondary fringes. In simple terms, the slit acts as a source of wavelets that interfere with each other.
Central maximum
The brightest, widest fringe at the centre of a single-slit diffraction pattern. It is twice as wide as any of the secondary maxima.
Dark fringe (single slit)
A position on the screen where light from the slit destructively interferes. Located at angles satisfying a sin θ = mλ, where m = ±1, ±2, ... (m = 0 is the central bright fringe, not a dark fringe).
The condition for bright fringes (maxima) is d sin θ = mλ, where d is the slit spacing, θ is the angle from the central axis, m is the order number (0, ±1, ±2, ...), and λ is the wavelength.
Slit spacing from line density: if a grating has 600 lines/mm, then d = 1/600 mm = 1.667 × 10⁻⁶ m.
The position of the m-th bright fringe on a screen at distance L is y_m = L tan θ_m. For small angles, y_m ≈ L sin θ_m, but at larger angles (common on exams) you must use the exact formula.
To find the spacing between two orders, calculate each angle separately and subtract the positions.
θ₁ = arcsin(mλ/d) for each order m
y₁ = L tan θ₁, y₂ = L tan θ₂
Spacing = y₂ − y₁
Grating: 600 lines/mm → d = 1/(600,000) m = 1.667 × 10⁻⁶ m
λ = 640.0 nm = 6.400 × 10⁻⁷ m
Screen distance L = 4.50 m
First order (m = 1): sin θ₁ = (1)(6.400 × 10⁻⁷) / (1.667 × 10⁻⁶) = 0.3839 → θ₁ = 22.60° → y₁ = 4.50 × tan(22.60°) = 1.871 m
Second order (m = 2): sin θ₂ = (2)(6.400 × 10⁻⁷) / (1.667 × 10⁻⁶) = 0.7678 → θ₂ = 50.15° → y₂ = 4.50 × tan(50.15°) = 5.393 m
Spacing = 5.393 − 1.871 = 3.52 m → Answer: C
Light hits the top surface of a thin film. Part reflects, part transmits into the film.
The transmitted part reflects off the bottom surface and comes back up.
These two reflected beams overlap and interfere.
The result depends on two things: the path difference (2t, where t is the film thickness, measured in the film's own wavelength λ_film = λ/n_film) and any phase shifts from reflection.
Reflection at a boundary where n increases (low to high): half-wavelength (π) phase shift.
Reflection at a boundary where n decreases (high to low): no phase shift.
If both reflections get a phase shift, or neither does, the shifts cancel and you only care about path difference.
If exactly one reflection gets a phase shift, there is a net half-wavelength shift that flips the interference condition.
Both reflections shift, or neither shifts (net zero phase shift):
Constructive: 2t = mλ_film (m = 0, 1, 2, ...)
Destructive: 2t = (m + 1/2)λ_film
Exactly one reflection shifts (net half-wavelength phase shift):
Constructive: 2t = (m + 1/2)λ_film
Destructive: 2t = mλ_film
Each object has three layers. Light reflects off the top and bottom surfaces of the middle layer (thickness d). The question asks which cases give constructive interference when d = λ_middle/4, i.e. 2d = λ_middle/2.
Case 1: layers 1.6 / 1.5 / 1.4. Top reflection: going from 1.6 to 1.5 (n decreases) → no shift. Bottom reflection: going from 1.5 to 1.4 (n decreases) → no shift. Net shift: zero. Constructive needs 2d = mλ_middle. Here 2d = λ_middle/2, which is not a whole number of λ_middle. Not constructive.
Case 2: layers 1.4 / 1.5 / 1.6. Top reflection: 1.4 to 1.5 (n increases) → shift. Bottom reflection: 1.5 to 1.6 (n increases) → shift. Net shift: zero (both shift, they cancel). Same condition as Case 1. Not constructive.
Case 3: layers 1.5 / 1.6 / 1.4. Top reflection: 1.5 to 1.6 (n increases) → shift. Bottom reflection: 1.6 to 1.4 (n decreases) → no shift. Net shift: half-wavelength. Constructive needs 2d = (m + 1/2)λ_middle. Here 2d = λ_middle/2 = (0 + 1/2)λ_middle. Constructive. Yes.
Case 4: layers 1.4 / 1.6 / 1.5. Top reflection: 1.4 to 1.6 (n increases) → shift. Bottom reflection: 1.6 to 1.5 (n decreases) → no shift. Net shift: half-wavelength. Same logic as Case 3. Constructive. Yes.
Answer: C (3 and 4)
Setup: air (n = 1.00) → MgF₂ coating (n = 1.38) → glass lens (n = 1.52).
Top reflection: air to MgF₂ (1.00 to 1.38, n increases) → phase shift.
Bottom reflection: MgF₂ to glass (1.38 to 1.52, n increases) → phase shift.
Both reflections shift → net phase shift is zero.
For destructive interference (the goal of an anti-reflective coating): 2t = (m + 1/2)λ_film.
Minimum thickness uses m = 0: 2t = λ_film/2 → t = λ_film/4.
λ_film = λ/n_film = 565/1.38 = 409.4 nm.
t = 409.4/4 = 102.3 nm ≈ 102 nm → Answer: A
For a single slit of width a, dark fringes appear at angles where a sin θ = mλ, with m = ±1, ±2, ±3, ... (m = 0 is not a dark fringe; it is the central maximum).
The central maximum extends from m = −1 to m = +1 and is twice the width of any secondary maximum.
The position of the m-th dark fringe on a screen at distance L is y_m = L tan θ_m.
For the first dark fringe (m = 1): sin θ = λ/a, then y = L tan θ.
λ = 696 nm = 6.96 × 10⁻⁷ m
Slit width a = 7.64 μm = 7.64 × 10⁻⁶ m
Screen distance L = 1.95 m
sin θ = λ/a = 6.96 × 10⁻⁷ / 7.64 × 10⁻⁶ = 0.09110
θ = arcsin(0.09110) = 5.228°
y = L tan θ = 1.95 × tan(5.228°) = 1.95 × 0.09148 = 0.1784 m = 17.8 cm → Answer: A
d \sin\theta = m\lambda \quad \text{(diffraction grating maxima)}a \sin\theta = m\lambda \quad \text{(single-slit dark fringes, } m = \pm1, \pm2, \ldots\text{)}\lambda_{\text{film}} = \frac{\lambda}{n_{\text{film}}}2t = m\lambda_{\text{film}} \quad \text{or} \quad 2t = \left(m + \tfrac{1}{2}\right)\lambda_{\text{film}} \quad \text{(thin-film interference; which is constructive depends on phase shifts)}y_m = L \tan\theta_m \quad \text{(position of fringe on screen)}Diffraction gratings are the core component in spectrometers, which split light from stars, chemical samples, or LEDs into their component wavelengths for analysis. Every time an astronomer identifies elements in a distant star, a grating is doing the work.
Anti-reflective coatings (thin-film interference) are on nearly every camera lens, pair of glasses, and smartphone screen. They reduce glare and increase the amount of light that passes through to the sensor or your eye.
Students often confuse the grating equation (d sin θ = mλ, for bright fringes) with the single-slit equation (a sin θ = mλ, for dark fringes). The same formula shape, but they describe opposite things: maxima for gratings, minima for single slits.
Students often forget to convert the wavelength to the wavelength inside the film (λ_film = λ/n) when solving thin-film problems. Using the vacuum wavelength directly gives the wrong thickness.
Students often assume that any reflection produces a phase shift. It does not. Only reflection at a boundary where the refractive index increases produces a half-wavelength shift.
Students sometimes use the small-angle approximation (sin θ ≈ tan θ) for diffraction grating problems where θ is large (above roughly 10°). At those angles the approximation breaks down and you must use exact trig.
⚠️ Thin-film interference problems almost always require you to check both reflection boundaries for phase shifts before choosing the constructive/destructive formula. This is the most commonly tested decision point.
⚠️ Diffraction grating problems at PHYS 116 level often involve angles large enough that you cannot approximate tan θ ≈ sin θ. Always compute θ with arcsin and then use L tan θ for the screen position.
⚠️ Anti-reflective coating questions test the minimum thickness, which corresponds to m = 0. Do not start at m = 1.
⚠️ Know the difference between d (grating spacing) and a (slit width). The exam may give line density (lines/mm), which you must convert to d in metres.
True or false: In single-slit diffraction, the first dark fringe corresponds to m = 0. (False. m = 0 is the central bright maximum. The first dark fringe is m = 1.)
Fill in the blank: A diffraction grating with 500 lines/mm has a slit spacing of ______ m. (2.00 × 10⁻⁶ m)
True or false: When light reflects from a low-n to high-n boundary, there is no phase shift. (False. Low-to-high gives a half-wavelength phase shift.)
Fill in the blank: The wavelength of 600 nm light inside a film of n = 1.5 is ______ nm. (400 nm)
True or false: For an anti-reflective coating where both reflections produce a phase shift, destructive interference requires 2t = (m + 1/2)λ_film. (True.)
Q: A diffraction grating with 400 lines/mm is illuminated with 550 nm light. What is the maximum order of bright fringe that can be observed?
A: d = 1/400,000 m = 2.50 × 10⁻⁶ m. The maximum order satisfies sin θ ≤ 1, so m_max = floor(d/λ) = floor(2.50 × 10⁻⁶ / 5.50 × 10⁻⁷) = floor(4.545) = 4. The highest observable order is m = 4.
Q: A thin film of oil (n = 1.48) floats on water (n = 1.33). What is the minimum thickness for constructive interference of 520 nm light viewed from above?
A: Top reflection (air to oil): n increases, so phase shift. Bottom reflection (oil to water): n decreases, so no phase shift. One shift only, so constructive needs 2t = (m + 1/2)λ_film. Minimum at m = 0: t = λ_film/4 = (520/1.48)/4 = 351.4/4 = 87.8 nm.
Q: Light passes through a 12.0 μm wide slit and falls on a screen 2.00 m away. If the distance from the central maximum to the first dark fringe is 10.5 cm, what is the wavelength of the light?
A: y = L tan θ → tan θ = 0.105/2.00 = 0.0525 → θ = 3.005°. a sin θ = λ → λ = 12.0 × 10⁻⁶ × sin(3.005°) = 12.0 × 10⁻⁶ × 0.05243 = 6.29 × 10⁻⁷ m = 629 nm.
Q: A soap film (n = 1.33) in air appears bright green (λ = 540 nm) in reflected light. What is the second smallest thickness that produces this?
A: Both reflections go from low to high n (air→soap and soap→air: wait, soap to air is high to low, so no shift there). Top: air to soap, shift. Bottom: soap to air, no shift. One shift, so constructive: 2t = (m + 1/2)λ_film. Second smallest is m = 1: 2t = (1.5)(540/1.33) = 1.5 × 406.0 = 609.0 nm → t = 304.5 nm.
Thin-film interference connects directly to Snell's law and the concept of refractive index from geometric optics. The refractive index determines both the wavelength inside the film and whether a phase shift occurs at each boundary.
Diffraction and interference also connect forward to topics in modern physics, where similar wave behaviour explains electron diffraction and the double-slit experiment with particles, which is foundational to quantum mechanics.
The single-slit diffraction pattern sets the resolution limit of optical instruments (the Rayleigh criterion), which you may encounter in the section on telescopes and microscopes.
Diffraction grating, grating equation, d sin theta = m lambda, order number, bright fringe, fringe spacing, thin film interference, thin-film, soap bubble colours, oil slick, anti-reflective coating, nonreflective coating, MgF2, magnesium fluoride, phase shift on reflection, half-wavelength shift, constructive interference, destructive interference, path difference, optical path length, wavelength in medium, lambda over n, single slit diffraction, single-slit, Fraunhofer diffraction, central maximum, dark fringe, a sin theta = m lambda, wave optics, physical optics, PHYS 116, Physics 116, UW, University of Washington, midterm 2, practice exam 2B