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In a Young's double-slit experiment, a detector at the centre of the fringe pattern reads an intensity I. If one of the two identical slits is now blocked, the intensity measured at that point becomes:

1. 2I
2. I
3. I/4
4. I/2

Correct answer: I/4

Solution

With both slits open, amplitudes add at the centre: total amplitude = 2a, so I = (2a)² = 4*I0 where I0 = a² is one slit's contribution. Blocking one slit leaves intensity I0 = I/4.

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