In an interference arrangement similar to Young's double slit experiment, the slits $$S1$$ and $$S2$$ are illuminated with coherent microwave sources each of frequency $$106Hz$$. The sources are synchronized to have zero phase difference. The slits are separated by distance d $$=$$ $$150$$ m. The intensity $$I($$θ$$)$$ is measured as a function of θ, where θ is defined as shown. If $$I0$$ is maximum intensity, then $$I($$θ$$)$$ for $$0$$≤θ≤$$90$$∘ is given by

Question

In an interference arrangement similar to Young's double slit experiment, the slits $$S1$$ and $$S2$$ are illuminated with coherent microwave sources each of frequency $$106Hz$$. The sources are synchronized to have zero phase difference. The slits are separated by distance d $$=$$ $$150$$ m. The intensity $$I($$θ$$)$$ is measured as a function of θ, where θ is defined as shown. If $$I0$$ is maximum intensity, then $$I($$θ$$)$$ for $$0$$≤θ≤$$90$$∘ is given by

Infinity Learn · Accepted Answer

For microwave λ$$=cf=3$$×$$108106=300m$$ As Δ$$x=dsin$$⁡θPhase difference λ$$=2$$πλ (Path$$ $$difference) ⇒λ$$=2$$πλ(dsin$$θ$$)=2$$$$π$$$$300(150sin$$θ$$)=$$λ$$sin$$θ$$IR=I1+I2+2I1I2cos$$ϕ Here $$I1=I2$$ and ϕ$$=$$π$$sin$$θ∴   $$IR=2I1$$[$$1+$$$$cos$$$$($$θ$$sin$$θ$$)]$$=4I1cos2$$$$π$$$$sin$$θ$$2IR$$ will be maximum when $$cos2$$$$π$$$$sin$$θ$$2=1$$∴ $$IRmax=4I1=Io$$ Hence $$I=I0cos2$$$$π$$$$sin$$θ$$2If$$ θ$$=0$$, then $$I=Iocos$$θ$$=IoIf$$ θ$$=30$$∘, then $$I=Iocos2($$π$$/4)=Io/2If$$ θ$$=90$$∘, then $$I=Iocos2($$π$$/2)=0$$

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