The total instantaneous electric field E at the point P on the screen is equal to the vector sum of the two sources:E=E+E.The magnitude of the Poynting vector flux S is proportional to the square of the total field. S∝E2=(E,+E2)2=E+E+2E,·E2 (14.2.10) Taking the time-average of S,the intensity of the light at P is I=(S)∝(E〉+(E)+2(E·E,〉 (14.2.11) The term 2EE2)represents the correlation between the two light waves.For incoherent light sources,since there is no definite phase relation between E and E2,the cross term vanishes,and the intensity due to the incoherent source is simply the sum of the two individual intensities, Line=+12. (14.2.12) For coherent sources,the term 2EE)is non-zero.In fact,for constructive interference,E=E,,and the resulting intensity is 1=411, (14.2.13) which is four times greater than the intensity due to a single source.When destructive interference takes place,E=-E2,andEE,and the total intensity becomes I=1-211+1=0, (14.2.14) as expected. Suppose that the waves emerged from the slits are coherent sinusoidal plane waves.Let the electric field components of the wave from slits 1 and 2 at P be given by E=E sin(ot), (14.2.15) and E2=E。sin(ot+φ)， (14.2.16) respectively,where the waves from both slits are assumed to have the same amplitude Eo. For simplicity,we have chosen the point P to be the origin,so that the kx dependence in the wave function is eliminated.Because the wave from slit 2 has traveled an extra 14-914-9 The total instantaneous electric field E  at the point P on the screen is equal to the vector sum of the two sources: E = E1 + E2    . The magnitude of the Poynting vector flux S is proportional to the square of the total field, 2 2 2 2 1 2 1 2 1 2 S ∝ E = (E + E ) = E + E + 2E ⋅E     . (14.2.10) Taking the time-average of S , the intensity I of the light at P is 2 2 1 2 1 2 I = S ∝ E + E + 2 E ⋅E   . (14.2.11) The term 1 2 2 E ⋅E   represents the correlation between the two light waves. For incoherent light sources, since there is no definite phase relation between E1  and E2  , the cross term vanishes, and the intensity due to the incoherent source is simply the sum of the two individual intensities, inc 1 2 I = I + I . (14.2.12) For coherent sources, the term 1 2 2 E ⋅E   is non-zero. In fact, for constructive interference, E1 = E2   , and the resulting intensity is 1 I = 4I , (14.2.13) which is four times greater than the intensity due to a single source. When destructive interference takes place, E1 = −E2   , and 1 2 1 E ⋅E ∝ −I   , and the total intensity becomes 1 1 1 I = I − 2I + I = 0 , (14.2.14) as expected. Suppose that the waves emerged from the slits are coherent sinusoidal plane waves. Let the electric field components of the wave from slits 1 and 2 at P be given by E1 = E0 sin(ωt) , (14.2.15) and 2 0 E = E sin(ωt +φ) , (14.2.16) respectively, where the waves from both slits are assumed to have the same amplitude E0 . For simplicity, we have chosen the point P to be the origin, so that the kx dependence in the wave function is eliminated. Because the wave from slit 2 has traveled an extra