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

Consider a two slit interference arrangements such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of $D$ in terms of $lambda$ such that the first minima on the screen falls at a distance $D$ from the centre $O$.

Solution:

From diagram

$T_1 P = T_1 O – O P = (D -x )$

$T_2 P = T_2 O + OP = (D + x )$

Now, $S_{1}P= sqrt{left(S_{1}T_{1}right)^{2}+left(T_{1}Pright)^{2}} $

$= sqrt{D^{2}+left(D-xright)^{2}} $

$ S_{2}P= sqrt{left(S_{2}T_{2}right)^{2}+left(T_{2}Pright)^{2}} $

$= sqrt{D^{2}+left(D+xright)^{2}}$

Path difference, $S_2P – S_1 P = frac{lambda}{2}$;

for first minimum to occur

$ sqrt{D^{2}+left(D+xright)^{2}}-sqrt{D^{2}+left(D-xright)^{2}} =frac{lambda}{2}$

The first minimum falls at a distance $D$ from the center, i.e.,

$x = D$.

$[D^2 + 4D^2]^{1/2} -D = frac{lambda}{2}$

$D(sqrt{5}-1) = frac{lambda}{2}$

$D( 2.236 -1) = frac{lambda}{2}$;

$D = frac{lambda}{2.472}$