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In Young's double-slit experiment, the widths of the two slits are in the ratio 4: 9. What is the ratio of the intensity at the bright fringes (maxima) to the intensity at the dark fringes (minima)?

1. 25: 1
2. 9: 4
3. 3: 2
4. 81: 16

Correct answer: 25: 1

Solution

Slit widths in ratio 4:9 => intensities I1:I2 = 4:9 (intensity proportional to width for incoherent sources). Amplitudes a1:a2 = sqrt(4):sqrt(9) = 2:3. At maxima (constructive interference): I_max proportional to (a1+a2)² = (2+3)² = 25. At minima (destructive interference): I_min proportional to (a1-a2)² = (3-2)² = 1. Ratio I_max: I_min = 25: 1.

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