In an interference arrangement similar to Young's double-slit experiment, the slits S_1 and S_2 are illuminated with coherent microwave sources, each of frequency 10^6 Hz. The sources are synchronized to have zero phase difference. The slits are separated by a distance d=150.0 m. The intensity I (theta) is measured as a function of theta, where theta is defined as shown. If I_0 is the maximum intensity, then I (theta) for 0lethetale90degree is given by

Options

A. (a) `I(theta)=4I_0` for `theta=0^@`
B. (b) `I(theta)=I_0//2` for `theta=30^@`
C. (c) `I(theta)=I_0//4` for `theta=90^@`
D. (d) `I(theta)` is constant for all values of `theta`

Answer

B

Explanation

For microwave `lambda=c/f=(3xx10^8)/(10^6)=300m`

As `Deltax=dsintheta`
Phase difference `lambda=(2pi)/(lambda)` (Path difference)
`implieslambda=(2pi)/(lambda)(d sin theta)=(2pi)/(300)(150sintheta)=lambdasintheta`
`I_R=I_1+I_2+2sqrt(I_1I_2)cosphi`
Here `I_1=I_2` and `phi=pisintheta`
`:. I_R=2I_1[1+cos(thetasintheta)]=4I_1cos2((pisintheta)/(2))`
`I_R` will be maximum when `cos^2((pisintheta)/(2))=1`
`:. (I_R)_(max)=4I_1=I_0`
Hence `I=I_0cos^2((pisintheta)/(2))`
If `theta=0`, then `I=I_ocostheta=I_o`
If `theta=30^@`, then `I=I_ocos^2(pi//4)=I//2`
If `theta=90^@`, then `I=I_ocos^2(pi//2)=0`