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In a Young's double-slit experiment, one slit is made twice as wide as the other (compared to the normal case of two equal-width slits). In the resulting interference pattern, which statement is correct?

1. The intensity of the maxima increases and the minima have nonzero (increased) intensity.
2. The intensities of both the maxima and the minima increase.
3. The intensity of the maxima increases and the minima have zero intensity.
4. The intensity of the maxima decreases and the minima have zero intensity.

Correct answer: The intensity of the maxima increases and the minima have nonzero (increased) intensity.

Solution

Making one slit wider lets more light through, so the amplitudes from the two slits are both larger than before AND now unequal (a1 != a2). Maxima Imax = (a1+a2)² increase because total amplitude is greater. Minima Imin = (a1-a2)² become nonzero (cannot fully cancel since amplitudes are unequal) and are larger than the original zero. So both maxima and minima intensities rise from their equal-slit values - maxima clearly increase, and the minima are no longer zero (they increase from zero).

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