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 = 150.0m`. The intensity `I(theta)` is measured as a function of `theta`, where `theta` is defined asa shown in fig. If `I_(0)` is hte maximum intensity, then `I(theta)` for `0 le theta le 90^(@)` is given by ltbrrgt

Options

A. `l(theta)=(l_(0))/(2) for theta=30^(@)`
B. `l(theta)=(l_(0))/(4) for theta=90^(@)`
C. `l (theta)=l_(0) for theta=0^(@)`
D. `l (theta)` is constant for all values of `theta.`

Answer

A, C

Explanation

for `theta=30^(@)`, DeltaP=d sin 30^(@) =(150 m)xx(1)/(2) =75 m.`
`rArr Deltaphi=75xx(2pi)/(lambda) =.^(75xx)(2pi)/(((3xx10^(8))/(10^(6))))=(pi)/(2)`
`rArr I_(res)= I+I+2I cos (Deltaphi) -2l`
where `I=(I_(0))/(4) {therefore "masimum intensity"=I_(0)=4I}rArr I_(res)=(I_(0))/(2)`
`theta=0^(@)` will automatically correspond to central maxima.
`theta=90^(@)` will correspond to path diff `d.=150 m ,rArr Deltaphi=((150xx2pi)/(3xx10^(8)))/(10^(6))=pi`
`rArr` minima
Hence intesity will be `0` at `theta=90^(@).`