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In a Young's double-slit experiment, the maximum intensity is I₀. The slit separation is d = 5*lambda, where lambda is the wavelength of monochromatic light. The screen is at a distance D = 10*d from the slits. Find the intensity of light at the point on the screen directly in front of one of the slits.

1. I₀
2. I₀ / 2
3. 3*I₀ / 4
4. I₀ / 4

Correct answer: I₀ / 2

Solution

The point directly in front of one slit is at transverse distance y = d/2 from the midpoint of the two slits. Path difference: Delta = y*d/D = (d/2)*(d)/(10d) = d/(20) = 5*lambda/20 = lambda/4. Phase difference: phi = 2*pi*Delta/lambda = 2*pi*(lambda/4)/lambda = pi/2. Intensity: I = I₀ * cos²(phi/2) = I₀ * cos²(pi/4) = I₀ * (1/sqrt(2))² = I₀/2.

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