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Two coherent sources emit light waves which superimpose at a point and are expressed as  E1=E0sin(ωt+π/4)E2=2E0sin(ωtπ/4)
Here, E1 and E2 are the electric field strengths of the two waves at the given point. If I is the intensity of wave expressed by field strength E1 . The resultant intensity istimes I.


Solution

Intensity of the wave expressed by field strength E1 is 
IE20As [intensity  (amplitude)2]
Intensity of the wave expressed by field strength E2 is
I(2E0)2    II=4 or   I=4I
where I is the intensity of wave expressed by field strength E1.
The phase difference between the two waves is given by
ϕ= (ωt+π/4)(ωtπ/4)=π2
Resultant intensity is given by  
IR=I+I'+2IIcosϕ(IR=I1+I2+2I1I2cosϕ]IR=I+4I+2I(4I)cosπ/2IR=5I
Thus, resultant intensity is five times the intensity of wave expressed by E1 

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