MCQ 14 Marks
In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes of $P_1$ and $P_2$ are orthogonal to each other. The polarizer $P_3$ covers both the slits with its transmission axis at $45^{\circ}$ to those of $P$ and $P_2$. An unpolarized light of wavelength $\lambda$ and intensity $I_0$ is incident on $P_1$ and $P_2$. The intensity at a point after $P_3$ where the path difference between the light waves from $s_1$ and $s_2$ is $\frac{\lambda}{3}$, is
- A$\frac{I_6}{2}$
- B$\frac{I_0}{4}$
- ✓$I_0$
- D$\frac{ I _0}{3}$
Answer
View full question & answer→Correct option: C.
$I_0$
(C) $I_0$
Sol.
after passing through third poleriser, Intensity of both the waves must be $\frac{I_0}{4}$
now, at a point where path diff is $\frac{\lambda}{3}$, phase difference$
\begin{array}{l}
\Delta \phi=2 K\left(\frac{\Delta x}{\lambda}\right)=\frac{2 \pi}{3} \\
\therefore I_{\text {ros }}=\sqrt{\left(\frac{I_0}{4}\right)^2+\left(\frac{I_0}{4}\right)^2+2\left(\frac{I_0}{4}\right)^2} \cos \frac{2 \pi}{3} \\
=\frac{I_0}{4}
\end{array}
$
Sol.
after passing through third poleriser, Intensity of both the waves must be $\frac{I_0}{4}$
now, at a point where path diff is $\frac{\lambda}{3}$, phase difference$
\begin{array}{l}
\Delta \phi=2 K\left(\frac{\Delta x}{\lambda}\right)=\frac{2 \pi}{3} \\
\therefore I_{\text {ros }}=\sqrt{\left(\frac{I_0}{4}\right)^2+\left(\frac{I_0}{4}\right)^2+2\left(\frac{I_0}{4}\right)^2} \cos \frac{2 \pi}{3} \\
=\frac{I_0}{4}
\end{array}
$
