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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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Answer» Solution :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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