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In a double slit experiment shown in figure, when light of wavelength 400 nm is used, dark fringe is observed at P. If D=0.2 m, the minimum distance between the slits S1 and S2 is mm. [figure]
0.20
Path difference at P (source → slit → P):
x= (SS1+S1P )- (SS2+S2P )
=2 D2+d2-2D
d D: D2+d2 D+ d22D x d2D
Dark fringe with minimum d: x= 2
d2= D2= 400 10-9 0.22=4 10-8 m2
d=2 10-4 m=0.20 mm
Answer: 0.20
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