Question

Consider the arrangement shown in figure (17-E4). The distance $D$ is large compared to the separation $d$ between the slits. (a) Find the minimum value of $d$ so that there is a dark fringe at $O$. (b) Suppose $d$ has this value. Find the distance $x$ at which the bright fringe is formed. (c) Find the fringe-width.

Answer 1

From the figure, $AB=BO$ and $AC=CO$

Path difference of the wave from reaching O,

$\Delta x=(AB+BO)-(AC+CO)$

$=2\left(A{B}_{A}C\right)$

$=2(\sqrt{D^2+d^2}-D)$

For dark fringe to be formed at O, path difference should be an odd multiple of $\frac{\lambda }{2}$

So,

$\Delta x=(2n+1)\frac{\lambda }{2}$

$2(\sqrt{D^2+d^2}-D)=(2n+1)\frac{\lambda }{2}$

$\sqrt{D^2+d^2}=D+(2n+1)\frac{\lambda }{4}$

$D^2+d^2=D^2+(2n+1{)}^2\frac{{\lambda }^2}{16}+(2n+1)\frac{\lambda D}{2}$

Neglecting $(2n+1{)}^2\frac{{\lambda }^2}{16}$ as it is very small, we get

$d=(\sqrt{2n}+1)\frac{\lambda D}{2}$

For minimum d putting n=0

$d_{min}=\sqrt{\frac{\lambda D}{2}}$

Thus for $d_{min}=\sqrt{\frac{\lambda D}{2}}$ there is a dark fringe at O.