The shift in the central fringe due to the introduction of a transparent sheet of thickness t and refractive index μ is given by Δy=dD(μ−1)t.
The position of the nth bright fringe is yn=ndDλ.
Equating the two, we get:
dD(μ−1)t=ndDλ
(μ−1)t=nλ
Substituting the given values:
(1.56−1)t=7×450×10−9
0.56t=3150×10−9
t=0.563150×10−9
t=5625×10−9 m
Comparing with t=α×10−9 m, we get α=5625.
Answer: 5625