Question

For a given YDSE experiment, $I_0$ is intensity due to each slit at screen. Slit separation is $d=0.2mm$. Distance of screen from slits is $D=1m$ and wavelength of light is $\lambda =400nm$.
If List - 1 gives distance of a point from center $C$ of the screen in $mm$ and List - 2 gives resultant intensity at that point, then match them for given YDSE.

List - 1	List - 2
(I) $y=0$	(P) 0
(II) $y=0.5$	(Q) $I_0$
(III) $y=1$	(R) $2I_0$
(IV) $y=\frac{4}{3}$	(S) $3I_0$
	(T) $4I_0$
	(U) None

• A. $I\to TII\to RIII\to PIV\to Q$
  
• B. $I\to TII\to SIII\to RIV\to Q$
• C. $I\to TII\to PIII\to QIV\to R$
• D. $I\to TII\to PIII\to SIV\to R$

Answer 1

The correct option is A $I\to TII\to RIII\to PIV\to Q$



Path difference,
$\Delta x=S_2{S}_2{}^{\prime }=S_2{S}_p-S_{1}P$
$\Delta x=dsin\theta =dtan\theta$
From diagram,
$tan\theta =\frac{y}{d}$
Hence,
$\Delta x=\frac{dy}{D}$
We know, $I_R=2I_0(1+cos\varphi )$
where $\varphi =\frac{2\pi }{\lambda }\Delta x$
When,
$y=0mm⟹\Delta x=0⟹\varphi =0⟹I_R=4I_0$
$y=0.5mm⟹\Delta x=100nm⟹\varphi =\frac{\pi }{2}⟹I_R=2I_0$
$y=1mm⟹\Delta x=200nm⟹\varphi =\pi ⟹I_R=0$
$y=\frac{4}{3}mm⟹\Delta x=\frac{800}{3}nm⟹\varphi =\frac{4\pi }{3}⟹I_R=I_0$