Question

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.

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	$\sqrt{\frac{\lambda D}{2}}$

• A. $A\to 4,B\to 3,C\to 2,D\to 1$
• B. $A\to 1,B\to 2,C\to 3,D\to 4$
• C. $A\to 1,B\to 3,C\to 2,D\to 4$
• D. $A\to 4,B\to 2,C\to 3,D\to 1$

Answer 1

Answer: (A)

Path difference, $P=\left(S{S}_2+S_{2}O\right)-\left(S{S}_1+S_{1}O\right)$
$S{S}_2=\sqrt{x^2+d^2}=x{\left(1+\frac{d^2}{x^2}\right)}^{1/2}=x\left(1+\frac{d^2}{2x^2}\right)$
$(\because d<<<x)$

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