A small transparent slab containing material of `mu=1.5` is placed along `AS_(2)`(figure). What will be the distance from O of the principle maxima and of the first minima on either side of the principal maxima obtained in the absence of the glass slab ?

Explanation

In case of transparent glass slabe of refractive indesx `mu`, the path difference will be calculated as `Deltax=2d sin theta+(mu-1)L`
In case of transparent glass slab of refractive index `mu`,
the path difference `=2d sin theta + (mu-1)L.`
For the principal maxima, (path difference is zero )
i.e., ` 2d sin theta _(0)+(mu-1)L=0`
or, `sin theta_(0)= -(L(mu-1))/(2d)=(-L(0.5))/(2d)" " [:. L=d//4]`
or `sin theta_(0)=(-6)/(16)`
`:. OP=D tan theta_(0)~~D sin theta_(0)=(-D)/(16)`
For the first minima, the path difference is `pm(lambda)/(2)`
`:. 2d sin theta_(1)+0.5L = pm(lambda)/(2)`
or `sin theta_(1)=(pm lambda//2 -0.5L)/(2d)=(pm lambda//2-d//8)/(2d)`
`=(pm lambda//2- lambda//8)/(2 lambda)=pm(1)/(4)-(1)/(16)`
[`:.` The diffraction occurs if the wavelength of waves in nearly equal to the side width (d)]
On the positive side `sin theta'_(1)^(+)= +(1)/(4)-(1)/(16)=(3)/(16)`
On the negative side `sin theta''_(1)^(-)=-(1)/(4)-(1)/(16)=-(5)/(16)`
The First principal maxima on the positive side is at distance
`D tan theta'_(1)^(+)=D ( sin theta'_(1)^(+))/(sqrt(1- sin^(2) theta'_(1)))=D(3)/(sqrt(16^(2)-3^(2)))=(3D)/(sqrt(247))` above point O
The first principal minima on the negative side is at distance
`D tan theta''_(1)=(5D)/(sqrt(16^(2)-5^(2))=(5D)/(sqrt(231))` below point O.