IIT JEE , AIEEE Forum - Physics , Optics

vpunithreddy

please provide me sollution for h.c verma problem no 16 in youngs double slit experiment

vpunithreddy

sir please reply this question

edison

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.