Here,since at the two slits we have introduced mica and polystyrene strips of

 

thickness 0.50mm = 0.05cm each with and refractive indices 1.58 and 1.55

 

respectively

 

₁ = 1.58

 

₂ = 1.55

 

So this will introduce extra path length of   times thickness

 

so extra path difference introduced is = (₁t -₂t) = (1.58 - 1.55) 0.05 cm

 

or extra path dfference = t = 0.0015cm

 

so in the expressions for young's double slite experiment we will introduce this

 

extra path difference to find fringe width and other parameters

 

so x  = dy/D + t 

 

here d = slit width = 0.12cm

 

 = 590nm

 

D = 1m = 100cm

 

so for constructive interference we use

 

x  = dy/D + t = n

 

and for destructive interference

 

x  = dy/D + t = n/2, where n is any integer

 

 Thus, introduction of 't' in the path lengh does not change the fringe width

 

as width = w = D/d = 100*590*10^-7/0.12 = 49.1 *10^-3cm = 4.91*10^-4cm

 

Now, to find the distance of first maximum from the center, we should

 

remember that the path difference at center is not zero but 't'.

 

so depending on  t = n or n/2 the central point will be maxima or minima or

 

neither of these.

 

Thus using above expressions and conditions we can find the distance of first

 

maximum from the center.