Difficulty: Intermediate | Prerequisites: PHYS 116 Midterm 1 material (wave basics, superposition), comfort with trigonometry and inverse trig functions.
This material sits at the heart of Midterm 2 for PHYS 116. You are moving from the basic wave model (amplitude, frequency, superposition) into what happens when light meets obstacles: slits, gratings, and thin films. The central question is always the same: how does the path-length difference between two or more waves determine whether they reinforce or cancel? If you are comfortable with the principle of superposition and can handle inverse trig on a calculator, you have what you need. Everything here connects back to the idea that waves arriving in phase add up (constructive interference) and waves arriving half a wavelength out of phase cancel (destructive interference).
Diffraction gratings and double slits produce bright fringes at angles where the path-length difference equals a whole number of wavelengths. Thin-film interference adds a twist: reflections at boundaries can introduce a half-wavelength phase shift, which flips the constructive and destructive conditions. Single-slit diffraction controls the width of the central bright band, and that width depends on wavelength, slit size, and distance to the screen.
Diffraction grating
An optical element with many equally spaced slits (often hundreds per millimetre) that splits light into sharp bright fringes at well-defined angles. Think of it as a much more precise version of a double slit, where the large number of slits makes the bright spots narrower and better separated.
Order number (m)
The integer labelling each bright fringe in a diffraction or interference pattern. 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 full wavelengths of path-length difference separate the waves arriving at that point.
Constructive interference
Occurs when two waves arrive in phase, so their amplitudes add. The path-length difference is a whole number of wavelengths (or an effective whole number after accounting for phase shifts). In simple terms, the peaks line up and the light is bright.
Destructive interference
Occurs when two waves arrive half a wavelength out of phase, so their amplitudes cancel. In simple terms, one wave's peak meets the other's trough, and you get darkness.
Thin-film interference
Interference between light reflected from the top and bottom surfaces of a thin transparent layer (soap film, oil slick, coating). The film's thickness, its refractive index, and any phase shifts from reflection together determine which wavelengths are reinforced or cancelled.
Phase shift on reflection
A half-wavelength (λ/2) shift that occurs when light reflects off a surface with a higher refractive index than the medium it is travelling through. Think of it as the wave flipping upside down on reflection. No shift occurs when reflecting off a medium with a lower refractive index.
Single-slit diffraction
The spreading of light as it passes through a single narrow opening, producing a broad central maximum flanked by weaker side fringes. In simple terms, the narrower the slit relative to the wavelength, the more the light fans out.
Central maximum width
The angular or linear width of the brightest band in a single-slit diffraction pattern. Formally given by w = 2λL/a for the linear width on a screen. In simple terms, it is the big bright stripe in the middle of the pattern.
The condition for bright fringes (constructive interference) is mλ = d sin θ, where d is the slit spacing, m is the order number, and λ is the wavelength.
To find d from a grating rated in lines/mm: d = 1 mm / (number of lines). For a 550 lines/mm grating, d = 1.82 × 10⁻⁶ m.
The position of the m-th bright fringe on a screen at distance L is y_m = L tan θ_m.
You cannot use the small-angle approximation (sin θ ≈ tan θ ≈ θ) for diffraction gratings, because the angles are typically too large. Always use the exact trig functions.
To find L given y and λ: solve for θ_m from the grating equation, then use L = y_m / tan θ_m.
When light reflects off a boundary where the second medium has a higher refractive index (n₂ > n₁), the reflected wave picks up a λ/2 phase shift.
When light reflects off a boundary where the second medium has a lower refractive index (n₂ < n₁), there is no phase shift.
Count the total number of phase shifts between the two interfering reflected rays. If only one ray undergoes a phase shift, the conditions flip relative to the "no phase shift" case.
Both rays shift, or neither shifts (even number of shifts):
Constructive: 2nt = mλ (m = 0, 1, 2, ...)
Destructive: 2nt = (m + 1/2)λ
Only one ray shifts (odd number of shifts):
Constructive: 2nt = (m + 1/2)λ
Destructive: 2nt = mλ
Here t is the film thickness, n is the film's refractive index, and λ is the wavelength in vacuum (or air).
To find which wavelengths in the visible spectrum (400 to 700 nm) satisfy the condition, solve λ = 2nt / (m + 1/2) or λ = 2nt / m for successive values of m and keep only those within the visible range.
Oil (n = 1.45) on water (n = 1.33): the top reflection (air to oil) shifts by λ/2 because oil has a higher index than air. The bottom reflection (oil to water) does not shift, because oil has a higher index than water. One shift total.
For destructive interference with one shift: 2nt = mλ.
The film thickness from drops: t = total volume / cross-sectional area = (number of drops × volume per drop) / A.
Longest wavelength corresponds to m = 1.
The central maximum has a linear width of w = 2λL / a, where a is the slit width, L is the slit-to-screen distance, and λ is the wavelength.
Proportionalities that matter: w is proportional to λ, proportional to L, and inversely proportional to a.
To widen the central maximum: increase λ, increase L, or decrease a.
To narrow the central maximum: decrease λ (e.g. submerge in water, which shortens the effective wavelength), decrease L, or increase a.
Submerging in water reduces the wavelength by a factor of n_water, thereby narrowing the pattern.
m\lambda = d \sin\theta_mGrating equation: relates order number m, wavelength λ, slit spacing d, and diffraction angle θ.
y_m = L \tan\theta_mPosition of the m-th bright fringe on a screen at distance L.
d = \frac{1 \text{ mm}}{N}Slit spacing from a grating with N lines per mm.
2nt = m\lambdaThin-film condition (even number of phase shifts): constructive when satisfied, destructive when 2nt = (m + 1/2)λ.
2nt = \left(m + \frac{1}{2}\right)\lambdaThin-film condition (odd number of phase shifts, i.e. one reflection shifts): constructive when satisfied, destructive when 2nt = mλ.
w = \frac{2\lambda L}{a}Linear width of the central maximum in single-slit diffraction.
Diffraction gratings are how spectrometers split light into its component wavelengths, which is the basis of chemical analysis, astronomy, and quality control in manufacturing. Thin-film interference is responsible for the colours you see in soap bubbles and oil slicks, and it is the principle behind anti-reflective coatings on camera lenses and glasses.
Students often use the small-angle approximation for diffraction gratings. The angles from a grating are large enough that sin θ and tan θ differ meaningfully, so always use exact trig.
Students frequently forget to check whether a phase shift occurs at each interface. You must compare refractive indices at every boundary, not just assume a shift happens or does not.
A common error is confusing which condition (mλ or (m + 1/2)λ) applies to constructive vs destructive interference. The answer depends entirely on the number of phase shifts. There is no single "constructive" formula.
Students sometimes assume the longest wavelength always corresponds to m = 0. For thin films with the (m + 1/2) condition, m = 0 gives the longest wavelength, but for the mλ condition, m = 1 gives the longest (m = 0 yields infinity, which is not physical).
⚠️ You will almost certainly be asked to determine which thin-film interference condition applies. Practise identifying the number of phase shifts quickly.
⚠️ Grating problems typically give lines/mm, not slit spacing. Convert first.
⚠️ Single-slit questions often ask which change widens or narrows the central maximum. Know the proportionalities cold: w ∝ λ, w ∝ L, w ∝ 1/a.
⚠️ Thin-film problems may ask for all visible wavelengths that satisfy a condition. You need to iterate through m values and filter for the 400 to 700 nm range.
True or false: A diffraction grating with more lines per mm produces a smaller slit spacing d. (True)
Fill in the blank: When light reflects off a medium with a ______ refractive index, it undergoes a λ/2 phase shift. (higher)
True or false: For a thin film in air where only one reflected ray is phase-shifted, constructive interference occurs when 2nt = mλ. (False, that is the destructive condition; constructive is 2nt = (m + 1/2)λ)
Fill in the blank: Decreasing the slit width in a single-slit experiment will ______ the central maximum. (widen)
True or false: You can safely use the small-angle approximation for a diffraction grating. (False)
Q: A 550 lines/mm diffraction grating is illuminated with 480 nm light. The second-order bright fringe lands 3.73 m from the central maximum on a distant screen. What is the grating-to-screen distance L?
A: First find d = 1 mm / 550 = 1.82 × 10⁻⁶ m. Then θ₂ = sin⁻¹(2 × 480 × 10⁻⁹ / 1.82 × 10⁻⁶). Finally L = 3.73 / tan θ₂ = 6.00 m.
Q: White light hits a soap film (n = 1.33, thickness 240 nm) in air at normal incidence. Which visible wavelengths are constructively reflected?
A: One phase shift (air-to-film boundary). Constructive: λ = 2nt / (m + 1/2). For m = 0: 1277 nm (infrared, not visible). For m = 1: 426 nm (visible). For m = 2: 255 nm (UV, not visible). Only 426 nm.
Q: Ten drops of oil (n = 1.45, each 5.00 × 10⁻¹⁰ m³) are placed on water (n = 1.33) in a dish of area 0.0314 m². What is the longest wavelength for destructive interference of the reflected light?
A: One phase shift (air-to-oil). Destructive: 2nt = mλ. Thickness t = 10 × 5.00 × 10⁻¹⁰ / 0.0314. Longest λ uses m = 1: λ = 2(1.45)(t) / 1 = 462 nm.
Q: A single-slit diffraction pattern shows a wide central maximum. Name two changes (made one at a time) that would widen it further.
A: Increase the wavelength λ, or decrease the slit width a. Either increases w = 2λL/a.
Thin-film interference connects directly to the wave model you studied in Midterm 1: path-length difference and superposition are the same principles, just applied to reflected rather than transmitted waves. Single-slit diffraction reappears when you study resolving power of telescopes and microscopes (Rayleigh criterion). The phase-shift rules at boundaries are the same physics behind anti-reflective coatings, which you may see again in engineering or materials courses.
diffraction grating, thin film interference, soap bubble colours, oil slick colours, constructive interference, destructive interference, phase shift on reflection, single slit diffraction, central maximum width, grating equation, path length difference, bright fringe, dark fringe, order number, anti-reflective coating, slit spacing, lines per mm, visible spectrum wavelengths, PHYS 116 midterm 2