Question

Two harmonic wave of monochromatic light

Y1= a cos ωt and Y2=a cos(ωt + Ф)

are superimposed on each other. Show that maximum intensity in interference pattern is four

times the intensity due to each split. Hence write the conditions for constructive and destructive

interference in terms of phase angle Ф.​

Answer 1

Solution :

By superposition principle, the resultant displacement at the observation point will be y=y1+y2

=y=y1+y2

=acoswt+acos(wt+ϕ)

=a[coswt+cos(wt+ϕ)]

=2acosϕ2.cos(wt+ϕ2)

Amplitude of resultant displacement =2acosϕ2

Since intensity α(amplitude)2

I=4ka2cm2ϕ2

Where k is the probability constant .

If I0 is the proportionality of each source

then I0=ka2 and

I=4I0cos2ϕ2

Hence for constructive interference :

cosϕ2=1

=> ϕ2=nπ

or ϕ=2nπ

For destructive interference

cosϕ2=0

=> ϕ2=(2n+1)π2

or ϕ=(2n+1)π