From the given data ,μ1 = 1.4 and μ2 = 1.7 . Let t be the thickness of each glass plates. Path difference at O, due to insertion of glass plates will be
Δx = ( μ2– μ1) t = (1.7 – 1.4) t = 0.3 t …(1) . The point to deduce here is that the path dfference would be lying between 5 λ and 5λ + λ/2 , since the 5th maxima and the 6th minima lies on both sides of O .
Hence , Δx is taken as 5λ + Δ …(2) . Here , Δ
Φ = 2π /λ . Δx = 2π /λ ( 5λ + Δ ) = ( 10 π + 2π/λ Δ ) …… (3)
Now , we know that I (Φ ) = Imax cos² [ Φ / 2 ] and the intensity at O is given as 3/4 I max . Therefore ,
3/4 I max = I (Φ ) = Imax cos² [ Φ / 2 ] or 3/4 = cos² [ Φ / 2 ] ….. (4)
Substituting the value of eq (3 ) in eq. (4) and solving , we get Δ = λ /6 .
Hence , Δx = 5λ + λ /6 = 31 λ / 6 . Now from eq. (1) Δx = 0.3 t . Therefore ,
t = 31 λ/6 x 0.3 = 9.3 x 10-6 m = 9.3 meu m