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In an interference arrangement similar t...


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 funtion of `theta`, where `theta` is defined as shown. if `I_(0)` is the maximum intensity then `I(theta)` for `0lethetale90^(@)` is given by:

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`

Text Solution

Verified by Experts

The correct Answer is:
AC

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)`
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