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In a Young's double-slit experiment one slit is wider than the other so that the amplitude of light from it is twice that from the other slit. If Iₘ is the maximum resultant intensity, the intensity I at a point where the phase difference is phi is:

1. (Iₘ/9)(1 + 8 cos²(phi/2))
2. (Iₘ/9)(4 + 5 cos(phi))
3. (Iₘ/3)(1 + 2 cos²(phi/2))
4. (Iₘ/5)(1 + 4 cos²(phi/2))

Correct answer: (Iₘ/9)(1 + 8 cos²(phi/2))

Solution

With amplitudes A and 2A, the resultant intensity I proportional to A² + 4A² + 4A² cos(phi) = 5A² + 4A² cos(phi). Using cos(phi) = 2cos²(phi/2) - 1: I proportional to A²(1 + 8 cos²(phi/2)). The maximum (phi=0) is Iₘ proportional to 9A². Dividing, I = (Iₘ/9)(1 + 8 cos²(phi/2)).

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