Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Consider the arrangement shows in the figure. The distance $D$ is large compared to the separation $d$ between the slits. For this arrangement, match the items in Column I with terms in Column II and choose the correct option from codes given below.
image
Column I Column II
A The minimum value of $d$ so that there is a dark fringe at $O$ for $x=D$ is 1 $\sqrt{\frac{\lambda D}{3}}$
B For $x=D$ and $d$ minimum such that there is dark fringe at $O$, the distance $y$ at which next bright fringe is located is 2 $2d$
C fringe width for $x=D$ 3 $d$
D The minimum value of $d$ so that there is a dark fringe at $O$ for $x=D / 2$ is 4 $\sqrt{\frac{\lambda D}{2}}$

Wave Optics

Solution:

Path difference, $P=\left(S S_{2}+S_{2} O\right)-\left(S S_{1}+S_{1} O\right)$
$S S_{2}=\sqrt{x^{2}+d^{2}}=x\left(1+\frac{d^{2}}{x^{2}}\right)^{1 / 2}=x\left(1+\frac{d^{2}}{2 x^{2}}\right)$
$(\because d< < < x)$
image
Similarly, $S_{2} O=\sqrt{\left(D^{2}+d^{2}\right)}=D\left(1+\frac{d^{2}}{2 D^{2}}\right)$
$(\because d < < D)$ Also, $\quad S S_{1} O=x+D$
Also, $S S_{1} O=x+D$
$\therefore P =x\left(1+\frac{d^{2}}{2 x^{2}}\right)+D\left(1+\frac{d^{2}}{2 D^{2}}\right)-(x+D)$
$=x+\frac{d^{2}}{2 x}+D+\frac{d^{2}}{2 D}-x-D$
or $P=\frac{d^{2}}{2}\left(\frac{1}{x}+\frac{1}{D}\right)$
For dark fringe, $P=\frac{\lambda}{2}$
[for minimum $d, P=\frac{(2 n-1) \lambda}{2} ; n=1$]
$\Rightarrow \frac{\lambda}{2}=\frac{d^{2}}{2}\left(\frac{1}{x}+\frac{1}{D}\right)$
or $d=\sqrt{\frac{\lambda x D}{(x+D)}}$
Put $x=D, d=\sqrt{\frac{\lambda D}{2}}$
$\Rightarrow$ Put $x=D / 2, d=\sqrt{\frac{\lambda D}{3}}$
Fringe width $=\beta=\frac{\lambda D}{d}=\frac{\lambda D}{\sqrt{\lambda D / 2}}=2 d$
Distance of next bright fringe from $O$.
Distance of consecutive bright and dark band
$=\frac{\text { Fringe width }}{2}=d$