Question

In a special arrangement of Young's double-slit experiment, the distance between the slits $d$ is twice the distance between the screen and the slits $D$ , i.e $d=2D$ . For this setup, the value of $D$ such that the first minima on the screen fall at a distance $D$ from the centre $O$ is found to be $\frac{\lambda }{N}$ .

What is the value of $N$ ? [Take $\sqrt{5}=2.24$ ]

Answer 1

Answer: 2.48

From diagram as provided in question, $OP=x$
$CO=D$
$S_{1}C=S_{2}C=D$ $T_{1}P=T_{1}O-OP=D-x$
$T_{2}P=T_{2}O+OP=D+x$
Now, $S_{1}p=\sqrt{{\left(S_1{T}_1\right)}^2+{\left(T_{1}P\right)}^2}=\sqrt{D^2+{\left(D-x\right)}^2}$
$S_{2}P=\sqrt{{\left(S_2{T}_2\right)}^2+{\left(T_{2}P\right)}^2}=\sqrt{D^2+{\left(D+x\right)}^2}$
For fist minimum to occur,
Path difference
$S_{2}P-S_{1}P=\frac{\lambda }{2}$
$\Rightarrow \sqrt{D^2+{\left(D+x\right)}^2}-\sqrt{D^2+{\left(D-x\right)}^2}=\frac{\lambda }{2}$
The first minimum falls at a distance $D$ from the centre, i.e., $x=D$
${\left[D^2+4D^2\right]}^{\frac{1}{2}}-D=\frac{\lambda }{2}$
$\therefore D\left(\sqrt{5}-1\right)=\frac{\lambda }{2}$
$\Rightarrow D=\frac{\lambda }{2\left(2.24-1\right)}=\frac{\lambda }{2.48}$