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Young's double-slit experiment gives you one main formula, β = λD/d, and a dozen ways to misuse it. Questions twist it with a liquid between the slits and screen, a thin sheet over one slit or slits of unequal brightness. This post works through each variation with numbers and points out the traps that appear year after year.
What does the experiment show, and what is the fringe width?
Light from two narrow, coherent slits a distance d apart overlaps on a screen a distance D away. Where the path difference is a whole number of wavelengths, the waves add and you see a bright fringe. Where it is an odd number of half-wavelengths, they cancel and you see a dark one. The NCERT Class 12 chapter Wave Optics derives the positions:
Bright fringes: x = nλD/d
Dark fringes: x = (n + ½)λD/d, for n = 0, ±1, ±2, …
Both are equally spaced, with fringe width β = λD/d. The JEE (Main) 2026 syllabus lists "Young's double-slit experiment and expression for fringe width", and the JEE (Advanced) 2026 brochure limits interference to Young's double-slit experiment. Check your own year's syllabus.
Why do students find interference confusing?
Ambrose, Shaffer, Steinberg and McDermott (1999) found that university students who had studied physical optics, at introductory level and beyond, often could not account for the pattern produced by a single or double slit. Many did not know whether to apply geometrical or physical optics and combined elements of both. In exam terms, that shows up as treating the slits like pinholes in ray diagrams, or expecting the fringes to bunch up away from the centre.
Worked example 1: fringe width and positions
Light of wavelength 600 nm falls on slits 0.20 mm apart, with the screen 1.0 m away.
β = λD/d = (600 × 10⁻⁹ × 1.0) / (0.20 × 10⁻³) = 3.0 × 10⁻³ m = 3.0 mm.
Third bright fringe from the centre: 3β = 9.0 mm.
Third dark fringe: n = 2 in the dark-fringe formula, so x = 2.5β = 7.5 mm.
Counting dark fringes is the most common slip. The first dark fringe is n = 0 at 0.5β, so the third is at 2.5β.
Worked example 2: the whole set-up in water
Immerse the same apparatus in water (refractive index 4/3). The wavelength in water is 600/(4/3) = 450 nm, so β = 450 × 10⁻⁹ × 1.0 / (0.20 × 10⁻³) = 2.25 mm. The fringes get closer together. A trap: if only the region between slits and screen were filled, the answer is the same, because that is where the path difference builds up.
Worked example 3: a thin film over one slit
A glass sheet of refractive index 1.5 and thickness 3.6 μm covers one slit. The sheet adds an extra optical path of (μ − 1)t, so the whole pattern shifts towards the covered slit.
Shift = (μ − 1)tD/d = (0.5 × 3.6 × 10⁻⁶ × 1.0) / (0.20 × 10⁻³) = 9.0 × 10⁻³ m = 9.0 mm.
In fringes: (μ − 1)t/λ = (0.5 × 3.6 × 10⁻⁶)/(600 × 10⁻⁹) = 3 fringes.
The fringe width does not change; only the position of the central maximum moves. Students often get the direction wrong: the centre moves towards the slit with the sheet, because light through that slit is slowed and the extra optical path has to be balanced by a shorter geometric path.
Worked example 4: unequal slits
If one slit lets through four times the intensity of the other (I₁ = I₀, I₂ = 4I₀), the resultant intensity is I = I₁ + I₂ + 2√(I₁I₂) cos φ. Maximum: (√I₀ + √4I₀)² = 9I₀. Minimum: (√4I₀ − √I₀)² = I₀. The ratio is 9:1, and the dark fringes are no longer completely dark.
Which other traps should you know?
Confusing slit separation with slit width. d is the distance between slit centres; slit width controls diffraction.
White light. The central fringe is white, because every wavelength has zero path difference there; the fringes either side are coloured, with red spaced wider than blue.
Units. Wavelengths come in nm or Å, separations in mm. Convert everything to metres first.
Moving the screen. β grows in proportion to D; the angular fringe width λ/d does not change.
Interference is far easier to understand once you can watch it. The Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read, so you can change d, D and λ and see the fringes respond. Our post on learning waves with a simulation suggests how to use one well.
A practice routine
Day 1: Five fringe-width and position questions, including at least two dark fringes.
Day 2: Three medium and thin-film questions, each with a sketch of which way the pattern moves.
Day 3: Two intensity questions with unequal slits.
Day 4: A timed mixed set of interference and single-slit diffraction questions.
Tonight, solve worked example 3 again with a sheet of index 1.6 and the same thickness, and check that your fringe shift comes out as 3.6.
