Intensity of Light at The Point of Interference - Wo06a

let’s start with the intensity of light at the point of interference. as we know, we already studied in the section of wave that, light intensity or intensity of any wave is directly proportional to square of its amplitude. for coherent waves we can consider the factor of frequency to be a constant. so intensity can be given as k ay square. and in case of interference we can write that, at the point of interference, if resulting amplitude is r, intensity at this point which can be written as, resulting intensity at the point is given as, here we can write resulting intensity as i r which is k r square, if we substitute the value of resulting amplitude we’ve calculated in the previous section. here i r can be given as k, ay 1 square plus, ay 2 square, plus 2 k ay 1 ay 2 coz phi. and here the value of resulting intensity at the point of interference is k ay 1 square plus, k ay 2 square, plus 2 k ay 1 ay 2 coz phi. here, in this expression we can see, the value k ay 1 square is the individual intensity of the 1st wave that we can write as i 1, this intensity of 2nd wave we can write as i 2, plus here k ay 1 ay 2 we can write as, twice of root i 1 i 2, coz phi. this is the expression we use for resulting intensity which depends on, individual intensities and a phase difference between the 2. so if we further analyze for constructive and destructive interferences we can write, for constructive interference, we know the value of coz phi is equal to plus 1, as the 2 waves are superposed in same phase, which implies, if coz phi is 1, here the resulting intensity is maximum and that will be i 1 plus i 2, plus, 2 root i 1 i 2. and here we can see the maximum intensity we’re getting is, by rearranging these terms, root i 1 plus root i 2 whole square. this is the intensity at the point of constructive interference, where the waves are super posed in same phase. similarly if we talk for destructive interference, where waves super pose in opposite phase the value of coz phi is equal to minus 1, which implies at the point of destructive interference intensity is minimum, which is given as i 1 plus i 2, minus 2 root of i 1 i 2. and here we can see by rearranging terms we get, root i 1 minus root i 2, whole square. so these maximum and minimum intensities are constructive and destructive interference points.