Given that, μ= 1.45, t = 0.02 mm = 0.02x10–3 m and λ= 620 nm = 620x10–9 m
We know, when the transparent paper is pasted in one of the slits, the optical path changes by (μ – 1)t.
Again, for shift of one fringe, the optical path should be changed by λ.
So, no. of fringes crossing through the centre is given by,