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Q.

In YDSE double slit experiment, one of the slit wider than other, so that the amplitude of the light from one slit is thrice of that from the other slit if I0 be the maximum intensity. Then the resultant intensity I when the interference at a phase difference  1200. is  xI016  then x is____  

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answer is 7.

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Detailed Solution

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Let  A1=A0,A2=3A0      ThenAmax2=A02+9A02+2A0.3A0=16A02        Amax=4A0...(i)        Forϕ=120       A2=A02+9A02+2A0×3A0×cos1200=10A023A02=7A02            IImax=110=7A0216A02        I=716I0

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