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Exam level9 past questions

Interference of light

Examine the principle of superposition, Young's double slit experiment, fringe width, intensity distribution, and the conditions for constructive and destructive interference.

Why this shows up in the exam

You must be able to solve problems on fringe patterns and intensity variations in NEET.

How NEET tests this

Numerical · 3 QsAssertion–reason · 2 QsDirect recall · 2 Qs

Learn the idea

Interference of light is the superposition of two or more coherent waves; the resulting bright or dark fringes are decided solely by the path difference between the waves – whole‑multiple of λ gives bright, half‑multiple gives dark.

🧠 Memory hook: Whole‑pizza slice = bright (whole λ), half‑slice bite = dark (half λ).

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • Constructive condition: path difference = m·λ (m=0,1,2…)
  • Destructive condition: path difference = (m+½)·λ
  • Fringe width β = λ·D / d for Young's double‑slit
  • Angular fringe width θ = λ / d (in radians)
  • Resultant intensity for equal amplitudes: I = 4I₀ cos²(δ/2)
  • For unequal intensities I₁,I₂: I_max = (√I₁+√I₂)² , I_min = (√I₁-√I₂)²

How to approach it

  1. 11. Note the wavelength(s) and whether the sources are coherent.
  2. 22. Write the geometric path difference (Δ = d·sinθ ≈ d·θ or Δ = d·x/D).
  3. 33. Apply the constructive (Δ=mλ) or destructive (Δ=(m+½)λ) condition to locate the required fringe or angular width.
  4. 44. Use β = λD/d (linear) or θ = λ/d (angular) for spacing; for intensity use the appropriate I‑formula.

Worked example — watch it click

If the monochromatic source in Young's double slit experiment is replaced by white light, then

  • ✅there will be a central bright white fringe surrounded by a few coloured fringes.
  • B)all bright fringes will be of equal width.
  • C)interference pattern will disappear.
  • D)there will be a central dark fringe surrounded by a few coloured fringes.

The concept behind this problem

The question asks what happens when many λ’s are present; it tests the insight that at zero path difference all colours add in phase (white centre) while away from centre each colour separates because β ∝ λ, producing coloured fringes.

Step by step

  1. 1In Young's double slit experiment with white light, all wavelengths interfere simultaneously.
  2. 2At the center (zero path difference), all wavelengths constructively interfere producing white light.
  3. 3Away from center, different wavelengths have maxima at different positions (since fringe width β = λD/d depends on λ).
  4. 4Red (longer λ) fringes are wider than violet.
  5. 5This produces a few colored fringes on either side of the central white fringe, after which overlap makes the pattern indistinct.

Watch out

Thinking that a white‑light source either gives only white fringes everywhere or erases the pattern altogether.

Common slip-ups that cost marks

  • •Treating white light as a single λ – different colours give different fringe spacings.
  • •Confusing angular width (λ/d) with linear width (λD/d).
  • •Missing a possible π phase change on reflection, which can shift the central bright to dark.

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Practise it

These are real questions from past NEET papers that test this exact idea.

Question 1 of 9NEET 2018

In Young's double slit experiment the separation d between the slits is 2 mm, the wavelength λ of the light used is 5896 Å and distance D between the screen and slits is 100 cm. It is found that the angular width of the fringes is 0.20°. To increase the fringe angular width to 0.21° (with same λ and D) the separation between the slits needs to be changed to:

Push further

More challenging

16 harder questions built from the past papers above — a step up in difficulty, with distractors designed so you can't get there by elimination. Written and checked by our reviewers, not from a real paper.

Question 1 of 16

In a Young's double-slit experiment, if the entire apparatus is immersed in water (refractive index n = 1.33) instead of air, how will the fringe width change?

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