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

Answer

AC

Explanation

The intensity of light is `I(theta)=I_(0)cos^(2)((delta)/(2))`
where `delta=(2pi)/(lamda)(Deltax)=((2pi)/(lamda))(dsintheta)`
(i) for `theta=30^(@)`
`lamda=(c)/(v)=(3xx10^(8))/(10^(6))=300m` and `d=150m`
`delta=((2pi)/(300))(150)((1)/(2))=(pi)/(2)therefore(delta)/(2)=(pi)/(4)`
`thereforeI(theta)=I_(0)cos^(2)((pi)/(4))=(I_(0))/(2)`
(ii). for `theta=90^(@)`
`delta=((2pi)/(300))(150)(1)=piimplies(delta)/(2)=(pi)/(2)` and `I(theta)=0`
(iii). for `theta=0^(@):delta=0implies(delta)/(2)=0thereforeI(theta)=I_(0)`