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Most Young's double-slit questions ask where the fringes are. A harder set asks how bright the light is at a given point, and that needs the intensity formula. Students often know I = 4I₀cos²(φ/2) by heart but cannot say where it comes from or why the average is not 4I₀. This post derives it, works through two problems, and covers the slips that cost marks.
Is interference intensity in the syllabus?
The Syllabus for JEE (Main) 2026 lists, under wave optics, "Young's double-slit experiment and expression for fringe width, coherent sources and sustained interference of light". The intensity result is Equation 10.11 in the NCERT Class 12 chapter on wave optics. Check the syllabus for your own exam year.
Where does I = 4I₀cos²(φ/2) come from?
NCERT's derivation takes two waves of the same amplitude a arriving at a point with phase difference φ:
y₁ = a cos ωt
y₂ = a cos (ωt + φ)
Adding them with the identity cos A + cos B = 2 cos((A - B)/2) cos((A + B)/2) gives
y = 2a cos(φ/2) cos(ωt + φ/2)
So the resultant is a wave of the same frequency with amplitude 2a cos(φ/2). Intensity is proportional to amplitude squared. One wave alone gives I₀ ∝ a², so the resultant gives
I = 4I₀cos²(φ/2)
The phase difference comes from the path difference Δ: φ = (2π/λ)Δ. Some values to remember:
φ = 0 (Δ = 0): I = 4I₀, a bright fringe
φ = π/2 (Δ = λ/4): I = 2I₀
φ = 2π/3 (Δ = λ/3): I = I₀
φ = π (Δ = λ/2): I = 0, a dark fringe
Why is the average intensity 2I₀ and not 4I₀?
The average of cos² over a full cycle is 1/2, so the average of 4I₀cos²(φ/2) across the fringes is 2I₀. That is exactly I₀ + I₀, the sum of the two separate intensities. Interference does not create energy; it moves it from dark fringes to bright ones.
If the two sources are not coherent, their phase difference changes rapidly and the pattern washes out. NCERT notes that you then see an average intensity of 2I₀ at every point (Equation 10.12). OpenStax University Physics defines coherent waves as those that are in phase or have a definite phase relationship, and incoherent waves as those with random phase relationships.
Worked example 1: path difference λ/3
This is NCERT Exercise 10.5. In a double-slit experiment with wavelength λ, the intensity where the path difference is λ is K units. Find the intensity where the path difference is λ/3.
Path difference λ: φ = 2π, so I = 4I₀cos²(π) = 4I₀. So 4I₀ = K.
Path difference λ/3: φ = (2π/λ)(λ/3) = 2π/3.
Intensity: I = 4I₀cos²(π/3) = 4I₀ × (1/2)² = I₀.
Answer: I = K/4.
Worked example 2: brightness at a point on the screen
Light of wavelength 600 nm falls on two slits 0.30 mm apart, with the screen 1.5 m away. Find the intensity 0.50 mm from the central bright fringe, as a fraction of the maximum.
Path difference. For small angles, Δ = yd/D = (0.50 × 10⁻³)(0.30 × 10⁻³) ÷ 1.5 = 1.0 × 10⁻⁷ m = 100 nm.
As a fraction of λ. 100 ÷ 600 = 1/6, so Δ = λ/6.
Phase. φ = 2π/6 = π/3.
Intensity. I = 4I₀cos²(π/6) = 4I₀ × 3/4 = 3I₀, which is three-quarters of the maximum 4I₀.
Check with the fringe width: β = λD/d = (600 × 10⁻⁹ × 1.5) ÷ (0.30 × 10⁻³) = 3.0 × 10⁻³ m = 3.0 mm. The point is at 0.50 mm, one-sixth of a fringe width from the centre, matching Δ = λ/6.
What if the two waves are not equally strong?
Adding two waves of amplitudes a₁ and a₂ with phase difference φ works like adding two vectors at angle φ. The resultant amplitude follows the cosine rule: A² = a₁² + a₂² + 2a₁a₂ cos φ. In intensities,
I = I₁ + I₂ + 2√(I₁I₂) cos φ
With amplitudes in the ratio 3 : 1, take a₁ = 3 and a₂ = 1. Then I_max = (3 + 1)² = 16 units and I_min = (3 - 1)² = 4 units, a ratio of 4 : 1. The dark fringes are no longer completely dark. Our post on Young's double-slit problems works another unequal-slit case along with fringe-shift questions.
Which mistakes cost marks?
Writing cos² φ instead of cos²(φ/2). At φ = π this gives 4I₀ instead of zero.
Mixing path difference and phase difference. Convert with φ = (2π/λ)Δ.
Using degrees in one step and radians in the next.
Adding intensities for coherent sources. That gives 2I₀ everywhere and no fringes.
Taking I₀ as the maximum. I₀ is one slit's intensity; the maximum is 4I₀.
A practice routine for this topic
Derive the formula from y₁ + y₂ once a week until you can do it in two minutes.
Make a table in your notebook of I for Δ = 0, λ/6, λ/4, λ/3 and λ/2, and check that the values fall from 4I₀ to zero.
Do one unequal-amplitude question each session, writing I_max and I_min as squares.
Mix in a fringe-position question so you practise turning a screen distance into a phase.
The rise and fall of brightness across fringes is easier to follow when you can watch it change. The Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read. For the history, read our story of Thomas Young and the wave theory of light.
Tonight, find the intensity where the path difference is λ/4 and confirm it is half the maximum.
