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In a Young’s double-slit setup, the intensity at a bright fringe maximum is I0. The slit separation is d = 5bb, where bb is the wavelength of the light used. If the screen is placed at a distance D = 10d, what intensity is observed at the point directly opposite one of the slits?

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

Correct answer: I0/2

Solution

Point opposite one slit is at y = d/2. Path difference = y*d/D = (d/2)(d)/(10d) = d/20 = (5*lambda)/20 = lambda/4. Phase = pi/2, so I = I0 cos^2(phase/2) = I0 cos^2(pi/4) = I0/2.

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