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⚛️ Physics  ·  Wave Optics  ·  NEET & JEE

In the double slit experiment, the intensity at a point where the path difference is λ/4 is (I<sub>0</sub> = intensity at central max):

Answer: I 0 /2.

  • A I<sub>0</sub>/2
  • B I<sub>0</sub>/4
  • C 0
  • D I<sub>0</sub>

Correct answer: A. I<sub>0</sub>/2

Explanation: Phase difference φ = 2π/λ × λ/4 = π/2. Intensity I = 4I<sub>1</sub> cos²(φ/2) = 4I<sub>1</sub> cos²(π/4) = 4I<sub>1</sub> × 1/2 = 2I<sub>1</sub>. Since I<sub>0</sub> = 4I<sub>1</sub>, I = I<sub>0</sub>/2.

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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