In a double slit experiment, two coherent sources have slightly different intensities `I` and `(I + delta I)`, such that `delta I lt lt I`. Show that resultant intensity at maxima is near `4 I`, while that at minima is nearly `(delta I)^(2)//4 I`.

In a double slit experiment, two coherent sources have slightly differ

1.	In a double slit experiment, two coherent sources have slightly different intensities `I` and `(I + delta I)`, such that `delta I lt lt I`. Show that resultant intensity at maxima is near `4 I`, while that at minima is nearly `(delta I)^(2)//4 I`.
Answer» From `I = I_(1) + I_(2) + 2sqrt(I_(1)I_(2)) cos phi` `I_(max) = I_(1) + I_(2) + 2sqrt(I_(1)I_(2)) cos 0^(@) = I + (I + delta I) + 2 sqrt(I(I + delta I))` As `deltaI lt lt I`, therefore, `I_(max) I + I 2 I = 4I` Again from `I = I_(1) + I_(2) + 2sqrt(I_(1)I_(2)) cos phi` `I_(min) = I + (I + deltaI) + 2sqrt(I(I + deltaI)) cos 180^(@) = 2I + delta I - 2 I(1 + (delta I)/(I))^(1//2)` `= 2 I + delta I - 2 I [1 + (1)/(2) (delta I)/(I) + ((1)/(2)((1)/(2) - 1))/(2!)((delta I)/(I))^(2)] = 2 I + delta I - 2 I - delta I + (1)/(4)I((delta I)/(I))^(2) = (delta I)^(2)/(4I)`

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In a double slit experiment, two coherent sources have slightly different intensities `I` and `(I + delta I)`, such that `delta I lt lt I`. Show that resultant intensity at maxima is near `4 I`, while that at minima is nearly `(delta I)^(2)//4 I`.

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