Questions
In an interference arrangement similar to Young's double slit experiment, the slits and are illuminated with coherent microwave sources each of frequency Hz. The sources are synchronized to have zero phase difference. The slits are separated by distance d = 150 m. The intensity I is measured as a function of q, where q is defined as shown. If is maximum intensity, then for is given by
detailed solution
Correct option is B
For microwave, λ=cf=3×108106=300 m Δx=dsinθ Using , ϕ=2πλΔx⇒ϕ=2πλ(dsinθ)=2π300(150sinθ)=πsinθIR=I1+I2+2I1I2cosϕPut I1=I2 and ϕ=πsinθ∴IR=2I1[1+cos(πsinθ)]=4I1cos2(πsinθ2)IR will be maximum when cos2(πsinθ2)=1∴(IR)max=4I1=IoHence, I=Iocos2(πsinθ2)If θ=30°, I=Iocos2(π/4)=Io/2If θ=60°, I=Iocos2(3π/4)If θ=90o, I=Iocos2(π/2)=0Talk to our academic expert!
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