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

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., $x=D$
$\left[D^{2} + 4 D^{2}\right]^{\frac{1}{2}}-D=\frac{\lambda }{2}$
$\therefore D\left(\sqrt{5} - 1\right)=\frac{\lambda }{2}$
$\Rightarrow D=\frac{\lambda }{2 \left(2 . 24 - 1\right)}=\frac{\lambda }{2 . 48}$