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(D) For microwaves,the wavelength is $\lambda = \frac{c}{f} = \frac{3 	imes 10^8}{10^6} = 300 \ m$.The path difference is $\Delta x = d \sin 	heta$.

Vedclass · Accepted Answer

Both $(a)$ and $(b)$

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(D) For microwaves,the wavelength is $\lambda = \frac{c}{f} = \frac{3 	imes 10^8}{10^6} = 300 \ m$.The path difference is $\Delta x = d \sin 	heta$.The phase difference is $\phi = \frac{2\pi}{\lambda} (\Delta x) = \frac{2\pi}{300} (150 \sin 	heta) = \pi \sin 	heta$.The resultant intensity is $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.Since $I_1 = I_2 = I$ (let),we have $I_R = 2I + 2I \cos \phi = 2I(1 + \cos \phi) = 4I \cos^2(\phi/2)$.Substituting $\phi = \pi \sin 	heta$,we get $I_R = 4I \cos^2\left(\frac{\pi \sin 	heta}{2}ight)$.The maximum intensity $I_0 = 4I$,so $I(	heta) = I_0 \cos^2\left(\frac{\pi \sin 	heta}{2}ight)$.For $	heta = 0^\circ$,$I(0) = I_0 \cos^2(0) = I_0$.For $	heta = 30^\circ$,$I(30^\circ) = I_0 \cos^2\left(\frac{\pi \sin 30^\circ}{2}ight) = I_0 \cos^2(\pi/4) = I_0 \left(\frac{1}{\sqrt{2}}ight)^2 = I_0/2$.For $	heta = 90^\circ$,$I(90^\circ) = I_0 \cos^2(\pi/2) = 0$.Thus,both $(a)$ and $(b)$ are correct.

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