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Interference fringes in Young's experiment have visibility (contrast). Visibility is maximum when:

Answer: Both slits have equal intensities.

  • A Both slits have equal intensities
  • B One slit is blocked
  • C The slits are very far apart
  • D The screen is very close

Correct answer: A. Both slits have equal intensities

Explanation: Visibility = (I<sub>max</sub> - I<sub>min</sub>)/(I<sub>max</sub> + I<sub>min</sub>). Maximum visibility = 1 occurs when both sources have equal intensity (I<sub>1</sub> = I<sub>2</sub>), giving I<sub>min</sub> = 0.

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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