## In This post you will get answer of Q.

In a YDSE, the light of wavelength l = 5000 $mathring{A}$ is

used, which emerges in phase from two slits a distance $d = 3 × 10^{-7}$m apart. A transparent sheet

of thickness $t = 1.5 × 10^{-7}m$ refractive index $mu = 1. 17$ is placed over one of the slits. what is the new angular position of the central maxima

of the interference pattern, from the centre of the screen? Find the value of y.

Solution:

## The path difference when transparent sheet is introduced $Delta x =(mu – 1) t$

If the central maxima occupies position of nth fringe ,then

$ (mu – 1) t = n lambda = d sin theta$

$Rightarrow :sin:theta :=:frac{left(mu :-:1right)t}{d}:=:frac{left(1.:17:-:1right)times 1.5:times 10^{-7}}{3times 10^{-7}}:=:0.085:$

$::Therefore,:angular:position:of:centralmaxima:$

$:theta :=:sin^{-1}:left(0.085right)=:4.88^{circ :}approx 4.9:$

$::For:For:small:angles,:sin:theta :approx theta :approx tan:theta :Rightarrow :tan:theta :=:frac{y}{D}:$

$:therefore :frac{y}{D}:=:frac{left(mu :-:1right)t}{d}:Rightarrow :y:=:frac{D:left(mu -:1right)t}{D} $