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Two coherent sources emit light waves which superimpose at a point where these can be expressed as E_1 = E_0sin (omega t + pi//4), E_2 = 2E_0) sin (omega t - pi//4)where E_1 and E_2are the electric strengths of the two waves at the given point. If I is the intensity of wave expressed by field strength E_1 , find the resultant intensity. |
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Answer» Solution :I being the intensity of wave expressed by field strength `E_1 , I alpha E_0^2`(intensity `alpha" (amplitude)"^2` ] Intensity of wave expressed by` E_2` `I. alpha (2E_0)^2` ` therefore (I.)/(I) =4, I.= 4I` Phase difference between the TWO waves` = (omegat+ pi//4) - (omega t -pi//4) = pi/2` Resultant intensity is given `I_R = I + I.+2 SQRT(II.) cos PHI` `therefore I_R = I + 4I = 2 sqrt(I(4I)) cos pi//2` `I_R = 5I` THUS resultant intensity is five times the intensity of wave expressed by `E_1` . |
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