Parallel monochromatic beam is falling normally on two slits S 1 and S 2 separated by d as 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 = 
, the intensity measured at P is Ι . and when z = 
, Intensity is x I. Find x.
Text Solution
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(2)
Sol. Where Z = 
= 
⇒ OS 4 = 
as shown.
If intensity at 'p' is Ι then intensity of light at S 3 and S 4 is Ι /4 & Ι /4
Path difference S 4 P – S 3 P = 0
So, intensity of slits S 1 and S 2
Δφ at S 4
Δ p = 
= 
= 
Δφ = 
= 
= Ι R = Ι 1 + Ι 1 + 
cos 
Ι 1 = Ι 2 = 
intensity of S 1 and S 2 .
If z = 
= 4 β
Δ p = 
= 
= 
= 2 λ
Δφ = 
.2d = 4 π
Ι 3 = Ι 4 = 
at 'p' Ι p = 
Ι p = 2 Ι .
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