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ExamsJEE MainPhysics

In Young's double-slit experiment the slit separation is d = 0.25 cm and the screen is at D = 100 cm. Light of wavelength lambda = 6000 angstrom is used, and I0 is the intensity at the central bright fringe. Find the intensity at a point x = 4 * 10⁻⁵ m from the central maximum.

  1. I0
  2. I0/2
  3. 3*I0/4
  4. I0/3

Correct answer: 3*I0/4

Solution

Path difference at x: delta = x*d/D = (4e-5)(0.25e-2)/(1.0) = 1e-7 m. Phase phi = (2*pi/lambda)*delta = (2*pi/6e-7)*(1e-7) = 2*pi/6 = pi/3. Intensity for two equal sources: I = I0*cos²(phi/2) = I0*cos²(pi/6) = I0*(sqrt(3)/2)² = 3*I0/4.

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