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Obtain the equaiton for resultant intensity due to interference of light. |
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Answer» Solution :Let us cosider two light waves from the sources `S_(1) and S_(2)` meeting at a point P. The wave from `S_(1)` at an INSTANT t at P is, `y_(1) = a_(1) sin omegat` The WAVR from `S_(2)` at an instant t at P is, `y_(2) = a_(2) sin (omegat + phi)` The two waves have different amplitudes `a_(1) and a_(2),` same angular frequency `omega` and a phase difference of `phi` between them. The resultant displacement will be given by, `y = y_(1) + y_(2) = a_(1) sin omega t + a_(1) (omega t + phi)` The simplification of the above equation by using trigonometric indentites gives the equation, `y = A sin (omega t + theta)` `"Where", A = sqrt(a_(1)^(2)+a_(2)^(2)+2a_(1)a_(2)cosphi)` `theta=tan^(-1)""(a_(2)sinphi)/(a_(1)+a_(2)cosphi)` The resltant amplitude is maximum, `A_(min)=sqrt((a_(1)+a_(2))^(2)),"when"phi=o,pm2pi,pm4pi...,` The resultant amplitude is minimum, `A_(min)=sqrt((a_(1)-a_(2))^(2)):"when"phi=pmpi,pm3pi,pm5pi...,` The intensity of light is prportional to square of amplitude, `1 PROP A^(2)` Now, equation (5) becomes, `1 prop I_(1) + I-(2) + 2 sqrt(I_(1)I_(2))cosphi` In equaiton (10) if the phase difference, `phi = 0, pm 2pi, pm4pi` .... it CORRESPONDS to the condition for maximum intensity of light called as constructive interference. The resultant maxiumum intensity is, `IpropI_(1)(a_(1)+a_(2))propI_(1)+I_(2)+2sqrt(I_(1)I_(2))` In equation (10) if the phase difference, `phi = pm pi, pm 3 pi, pm 5 pi`... it corresponds to the condition for minimum intensity of light called destructive interference. The resultant minimum intensity is, `I_(min), prop (a_(1) - a_(2))^(2) prop I_(1) + I_(2) - 2 sqrt(I_(1)I_(2))` 
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