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⚛️ Physics  ·  Wave Optics  ·  NEET & JEE

Light of wavelengths 400 nm and 600 nm is used in YDSE. What is the minimum distance from centre where fringes coincide?

Answer: 6β for 400 nm.

  • A 6β_600 = 4β_600 = bright fringe at 3 mm if D=1m, d=0.2mm
  • B 4β for 600 nm
  • C 6β for 400 nm
  • D 12β for both

Correct answer: C. 6β for 400 nm

Explanation: n<sub>1</sub> λ_1 = n<sub>2</sub> λ_2. 400 n<sub>1</sub> = 600 n<sub>2</sub>. Minimum: n<sub>1</sub> = 3, n<sub>2</sub> = 2. Position = n<sub>1</sub> λ_1 D/d = 3 × 400 nm × D/d.

Young's Double Slit Experimentsourceslits S₁,S₂dscreenbrightdarkPath difference at the screen determines bright (constructive, nλ) vs dark (destructive, (2n-1)λ/2) fringes

Two coherent slits S₁ and S₂ act as secondary sources; at each point on the screen, the path difference between light from S₁ and S₂ determines whether the waves arrive in phase (bright fringe) or out of phase (dark fringe), producing the characteristic alternating fringe pattern.

Concept context

Huygens principle, Young's double slit, diffraction, polarization. Essential for JEE.

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