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In a Young's double-slit experiment, D is the screen-to-slit distance and d is the slit separation. What is the distance from the central maximum to the nearest point on the screen where the intensity equals the intensity that one slit alone would produce?

1. D*lambda/d
2. D*lambda/(2*d)
3. D*lambda/(3*d)
4. 2*D*lambda/d

Correct answer: D*lambda/(3*d)

Solution

With each slit alone giving intensity I0, the combined pattern is I = 4*I0*cos²(phi/2), where phi is the phase difference. Requiring I = I0 gives cos²(phi/2) = 1/4, so phi/2 = pi/3 and phi = 2*pi/3. This corresponds to a path difference delta = (lambda/2*pi)*phi = lambda/3. Using delta = y*d/D, the position is y = D*delta/d = D*lambda/(3*d). This is the nearest such point.

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