Optics Ques 148
- Column I shows four situations of standard Young’s double slit arrangement with the screen placed far away from the slits $S _1 \quad$ 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 $S _1 P _2-S _2 P _2=\lambda / 2$, where $\lambda$ is the wavelength of the light used. In the cases $B, C$ and $D$, 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.
(2009)
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Answer:
Correct Answer: 148.$ (A) → p, s; (B) → q; (C) → t; (D) → r, s, t$
Solution:
Formula:
- (A) $\rightarrow(p, s) \rightarrow$ 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, $\delta\left(P _1\right)=0$.
(C) $\rightarrow(t) \rightarrow \delta\left(P _0\right)=\frac{\lambda}{2}, \delta\left(P _1\right)=\frac{\lambda}{2}-\frac{\lambda}{4}=\frac{\lambda}{4}$ 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})>I\left(P _0\right)\right. $
(D) $\rightarrow(r, s, t)$,
$ \begin{aligned} \delta\left(P _0\right) & =\frac{3 \lambda}{4}, \delta\left(P _1\right)=\frac{3 \lambda}{4}-\frac{\lambda}{4}=\frac{\lambda}{2} \\ \text { and } \quad \delta\left(P _2\right) & =\frac{3 \lambda}{4}-\frac{\lambda}{3}=\frac{5 \lambda}{12} . \end{aligned} $
In this case $I\left(P _1\right)=0$.
BETA
