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In Young's double slit experiment, one of the slit is wider than other, so that amplitude of light from one slit is double that of other slit. If Iₘ be the maximum intensity, the resultant intensity I when they interfere at phase difference given by:

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

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

Solution

The correct option accounts for the amplitude difference between the two slits, leading to a resultant intensity that incorporates the phase difference and the squared amplitudes. The factor of 8 in the cosine term arises from the specific ratio of the amplitudes, reflecting how the interference pattern is modified due to the unequal slit widths.

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