Question

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

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

If slit $S_1$ and slit $S_2$ are covered by thin films each of thickness $2000nm$ and of refractive index 1.5 and 1.2 respectively, then match List - I with List - II

• A. $I\to TII\to SIII\to RIV\to P$
• B. $I\to PII\to SIII\to RIV\to T$
• C. $I\to PII\to RIII\to TIV\to R$
• D. $I\to TII\to QIII\to PIV\to R$

Answer 1

The correct option is C $I\to PII\to RIII\to TIV\to R$

Now path difference becomes,
$\Delta x=\left[\right(S_{2}P-t)+{\mu }_{2}t]-\left[\right(S_{1}P-t)+{\mu }_{1}t]$
$\Delta x=(S_{2}P-S_{1}P)+({\mu }_2-{\mu }_1)t$
$=\frac{dy}{D}+({\mu }_2-{\mu }_1)t$
$=\frac{dy}{D}+600nm$
and $I_R=2I_0(1+cos\varphi )$
where $\varphi =\frac{2\pi }{\lambda }\Delta x$
When,
$y=0mm⟹\Delta x=600nm⟹\varphi =3\pi ⟹I_R=0$
$y=0.5mm⟹\Delta x=700nm⟹\varphi =\frac{7\pi }{2}⟹I_R=2I_0$
$y=1mm⟹\Delta x=800nm⟹\varphi =4\pi ⟹I_R=4I_0$
$y=\frac{4}{3}mm⟹\Delta x=\frac{2600}{3}nm⟹\varphi =\frac{13\pi }{2}⟹I_R=2I_0$