Wave OpticsMedium
Question
In Young′s double slit experiment with light of wavelength λ = 600 nm, intensity of central fringe is I0. Now one of the slit is covered by glass plate of refraction index 1.4 and thickness t = 5 μ m, the new intensity at the same point on screen will be :-
Options
A.
B.
C.I0
D.
Solution
Optical path diff ᐃx = (μ - 1)t
Phase diff ᐃφ =
ᐃx = 
× 0.4 × 5 × 10-6
ᐃφ =
Ires = I0cos2
- I0cos2
Phase diff ᐃφ =
ᐃx = × 0.4 × 5 × 10-6ᐃφ =
Ires = I0cos2
- I0cos2Create a free account to view solution
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