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_1 S_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 ` .

Answer

A

Explanation

`|y_(S_3)| = |y_(S_4)| = z/2 = y` (say)
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` .