Question

Two coherent sources emit light waves which superimpose at a point and are expressed as E1=E0sin(ωt+π/4)E2=2E0sin(ωt−π/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 is times I.

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 is

Solution

Intensity of the wave expressed by field strength E1 is

I∝E20As [intensity ∝ (amplitude)2]

Intensity of the wave expressed by field strength E2 is

I′∝(2E0)2 I′I=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)−(ωt−π/4)=π2

∴ Resultant intensity is given by

IR=I+I'+2√II′cosϕ⇒(IR=I1+I2+2√I1I2cosϕ]⇒IR=I+4I+2√I(4I)cosπ/2⇒IR=5I

Thus, resultant intensity is five times the intensity of wave expressed by E1

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