Q48 / 57 · JEE Advanced 2009 (Paper 2) · Physics

Column I shows four situations of standard Young's double slit arrangement with the screen placed far away from the slits $S_1$ and $S_2$. In each of these cases $S_1 P_0=S_2 P_0$, $S_1 P_1-S_2 P_1=\frac{\lambda}{4}$ and $S_1 P_2-S_2 P_2=\frac{\lambda}{3}$, where $\lambda$ is the wavelength of the light used. In the cases $\mathrm{B}, \mathrm{C}$ and $\mathrm{D}$, a transparent sheet of refractive index $\mu$ and thickness $t$ is pasted on slit $S_2$. The thickness of the sheets are different in different cases. The phase difference between the light waves reaching a point $P$ on the screen from the two slits is denoted by $\delta(P)$ and the intensity by $I(P)$. Match each situation given in Column-I with the statement(s) in Column-II valid for that situation. Figure

  • A
    (A) q,t, (B) q, (C) t, (D) r,s,t
  • B
    (A) q,r,s, (B) r, (C) q,r, (D) q,s,t
  • C
    (A) p,s, (B) r, (C) q,t, (D) r,s
  • Correct answer
    D
    (A) p,s, (B) q, (C) t, (D) r,s,t

Solution

(A) $\rightarrow(\mathrm{p}, \mathrm{s}) \rightarrow$ Intensity at $P_0$ is maximum. It will continuously decrease from $P_0$ towards $P_2$. $(B) \rightarrow(q) \rightarrow$ Path difference due to slap will be compensated by geometrical path difference. Hence, $\delta\left(P_1\right)=0$. $(\mathrm{C}) \rightarrow(\mathrm{t}) \rightarrow \delta\left(P_0\right)=\frac{\lambda}{2}, \delta\left(P_1\right)=\frac{\lambda}{2}-\frac{\lambda}{4}=\frac{\lambda}{4}$ and $\delta\left(P_2\right)=\frac{\lambda}{2}-\frac{\lambda}{3}=\frac{\lambda}{6}$. When path difference increases from 0 to $\frac{\lambda}{2}$, intensity will decrease from maximum to zero. Hence, in this case, $$ I\left(P_2\right)>I\left(P_1\right)>I\left(P_0\right) $$ $(\mathrm{D}) \rightarrow(\mathrm{r}, \mathrm{s}, \mathrm{t})$ $$ \delta\left(P_0\right)=\frac{3 \lambda}{4}, \delta\left(P_1\right)=\frac{3 \lambda}{4}-\frac{\lambda}{4}=\frac{\lambda}{2} $$ and $\delta\left(P_2\right)=\frac{3 \lambda}{4}-\frac{\lambda}{3}=\frac{5 \lambda}{12}$ In this case $I\left(P_1\right)=0$

Hence, the correct answer is option: 'D'.

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