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Consider the arrangement shown in figure. By some mechanism, the separation between the slits S_3 and S_4 can be changed. The intensity is measured at the point P which is at the common perpendicular bisector of S_1S_2 and S_3 S_4. When z=(Dlambda)/(2d), the intensity measured at P is I. Find the intensity when z is equal to (a)(Dlambda)/d(b)(3Dlambda)/(2d)(c) (2Dlambda)/d. |
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Answer» When, `z=(Dlambda)/(2D), z/2=y =(Dlambda)/(4d) ` `:.Deltax = (yd)/D =lambda/4` and we have seen in the above example that, at `Deltax = lambda/4`, intensity is `2I_0`. `:.I_S_3 = I_S_4 = 2I_0` Now, P is at the perpendicular bisector of `S_3 S_4`. Therefore, intensity at P will be four times of `2I_0` or `8I_0`. `8I_0 =I` Hence, `I_0 =I/8` (a)When `z=(Dlambda)/d` `y=z/2 = (Dlambda)/(2d)` ` :.Deltax = (yd)/D = lambda/2` or` I_S_(3) = I_S_(4)=0` Hence, ` I_p=0` (b) When ` z = (3Dlambda)/(2d)` ` y=z/2 =(3Dlambda/4d)` `Delta x = (yd)/D = (3lambda)/4` `:. Deltaphi or phi = 2pi/lambda(Deltax) = 3pi/2` Using`I=4I_0cos^2phi/2` We have, `I_(S_3) =I_(S_4)=2I_0 ` ` :. I_p = 4(2I_0) = 8I_0 = I` (c) When `z=(2Dlambda)/d` `y = z/2 = (Dlambda)/d` `:. Delta x = yd/D = lambda` `:. I_(S_3) = I_(S_4) = 4I_0 ` ` I_p = 4(4I_0) = 16I_0 = 2I` . |
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