Question

Solution. By the principle of superposition, the resultant displacement at the observation point will be
[
begin{aligned}
y &=y_{1}+y_{2}=a[cos omega t+cos (omega t+phi)]
&=2 a cos frac{phi}{2} cdot cos left(omega t+frac{phi}{2}right)
end{aligned}
]
Amplitude of the resultant displacement ( =2 a cos frac{phi}{2} )
( because ) Intensity ( propto(text { amplitude })^{2} )
( therefore ) Intensity, ( quad I=4 k a^{2} cos frac{phi}{2} )
where ( k= ) a proportionality constant. If ( I_{0} ) is the intensity of each source, then ( I_{0}=k a^{2} ) and
[
I=4 I_{0} cos ^{2} frac{phi}{2}
]
For constructive interference :
[
cos frac{phi}{2}=pm 1 text { or } frac{phi}{2}=n pi text { or } phi=2 n pi
]
mathrm{For destructive interference : } ~
[
cos frac{phi}{2}=0 text { or } frac{phi}{2}=(2 n+1) frac{pi}{2} text { or } phi=(2 n+1) pi
]

# Problem 5. Two monochromatic waves emanating from two coherent sources have the displacements represented by y, = a cosot, and Y2 = a cos(ot +), where 0 is the phase difference between the two displacements. Show that the resultant intensity at a v) point due to their superposition is given by I = 41, cos? 0/2, where lo oca?. Hence obtain the e conditions for constructive and destructive interference. le

Solution