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 $$106$$ Hz. 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 q, where q is defined as shown. If $$I0$$ is maximum intensity, then $$I($$θ$$)$$ for $$0$$≤θ≤$$90o$$ 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 $$106$$ Hz. 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 q, where q is defined as shown. If $$I0$$ is maximum intensity, then $$I($$θ$$)$$ for $$0$$≤θ≤$$90o$$ is given by

Infinity Learn · Accepted Answer

For microwave,  λ$$=cf=3$$×$$108106=300$$ m Δ$$x=dsin$$θ Using , ϕ$$=2$$πλΔx⇒ϕ$$=2$$πλ(dsin$$θ$$)=2$$$$π$$$$300(150sin$$θ$$)=$$π$$sin$$θ$$IR=I1+I2+2I1I2cos$$ϕPut $$I1=I2$$ and ϕ$$=$$π$$sin$$θ∴$$IR=2I1$$[$$1+$$$$cos$$$$($$π$$sin$$θ$$)]$$=4I1cos2($$π$$sin$$θ$$2)IR$$ will be maximum when $$cos2($$π$$sin$$θ$$2)=1$$∴$$(IR)max=4I1=IoHence$$, $$I=Iocos2($$π$$sin$$θ$$2)If$$ θ$$=30$$°, $$I=Iocos2($$π$$/4)=Io/2If$$ θ$$=60$$°, $$I=Iocos2(3$$π$$/4)If$$ θ$$=90o$$, $$I=Iocos2($$π$$/2)=0$$

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