Question

In the YDSE, the monochromatic source of wavelength `lambda` is placed at a distance `d/2` from the central axis (as shown in the figure), where d is the separation between the two slits `S_1 and S_2` .

(a)Find the position of the central maxima.
(b) Find the order of interference formed at O.
(c)Now, S is placed on centre dotted line. Find the minimum thickness of the film of refractive indes `mu =1.5` to be placed in front of `S_2` so that intensity at O becomes `3/4`th of the maximum intensity.
(Take `lambda=6000Å, d = 6mm`.)

Answer 1

(a) `(Deltax)_"net" = 0`
`:. (y_1d)/D_1 = (y_2d)/D_2`
` :. (d//2)/(1.5) = y/(2.0)`
or `y= d/1.5`
`=6/1.5`
=4mm.
(b) At O, net path difference,
`Deltax = (y_1d)/D_1`
`= (d//2)(d)/D_1`
` =(6xx 10^(-3))^2)/(2 xx 1.5)`
`= 12 xx 10^(-6)m `
` = 120 xx 10^(-7) m`
` lambda=6000Å`
`= 6 xx 10^(-7)`m
As, `Deltax = 20lambda`, therefore at O bright fringe of
order 20 will be obtained.
`(c) I = I_(max) "cos"^2phi/2`
`3/4I_(max) = I_(max) "cos"^2 phi/2`
`:. phi/2 = pi/6`
`phi= pi/3 = (2pi)/lambda(mu-1)t`
`:. t=lambda/(6(mu-1))`
`6000/(6(1.5-1))`
=2000Å.