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| 1. |
Obtain the expression for fringe width in the case of interference of light waves. |
Answer» Solution : Consider two coherent sources `S_1` and `S_2`separated by a distance .d.. Let a screen be placed at a distance .D. from the coherent sources. The point .O. on the screen is equidistant from `S_1` and `S_2` . So that path difference between the two light waves from `S_1` and `S_2`reaching O is zero. Thus the point o has maximum INTENSITY. Consider point .P. at a distance .x. from O.. The path difference between the light waves from `S_1` and `S_2`reaching point P is `DELTA = S_2P - S_1P` From the figure `Delta^("le") S_2 PF` `(S_2P)^2 = (S_2F)^2 + (FP)^2` ` = D^2 + (x + d/2)^2` Similarly `Delta^("le")S_1PE` `(S_1P)^2 = (S_1E)^2 + (EP)^2` `=D^2 + (x- d/2)^2` `(S_2P)^2 - (S_1P)^2 = [D^2 + (x + d/2)^2] - [D^2 + (x - d/2)^2] = [D^2 + x^2 + (d^2)/(4) + 2(x) (d/2)] - [D^2 + x^2 + (d^2)/(4) - 2(x) (d/2) ]` `(S_2P)^2 - (S_1P)^2- 2xd` or `S_2P - S_1P = (2xd)/((S_2P + S_1P))` Since .P. is very close to O and d < < D `S_2P + S_1P ~~2D` `therefore` Path difference : `S_2P - S_1P = (2xd)/(2D) = (xd)/(D) to (1)` Equation (1) represents the path difference between light waves from `S_1` and `S_2`superposing at point P. For bright fringe or maximum intensity at P. `S_2P - S_1P = n lamda"" n = 0,1,2,3.......` From eqn(1)`(xd)/(D) = n lamda` or `x = (n lamdaD)/(d)` The distance of the nth bright fringe from the centre .O. of the screen is `x_n = n ((lamda D)/(d))` Thedistance of the `(n + 1)^(th)`bright fringe from the centre .O. of the screen is `x_(n+1) = (n+1) ((lamda D)/(d))` By definition of fringe width `beta = x_(n+1) - x_n` ` = (n+1) ((lamda D)/(d)) - n ((lamda D)/(d))` `(lamda D)/(d) (n +1 -n)` `beta =(lamda D)/(d)` |
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