Question 84: Waves Optics

Question

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 $P_1$ 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

Accepted Answer

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
$$\begin{aligned}
& \Delta \phi=2 \mathrm{~K}\left(\frac{\Delta \mathrm{x}}{\lambda}ight)=\frac{2 \pi}{3} \
& 	herefore \mathrm{I}_{	ext {res }}=\sqrt{\left(\frac{\mathrm{I}_0}{4}ight)^2+\left(\frac{\mathrm{I}_0}{4}ight)^2+2\left(\frac{\mathrm{I}_0}{4}ight)^2} \cos \frac{2 \pi}{3} \
& =\frac{\mathrm{I}_0}{4}
\end{aligned}$$

Answer

$\mathrm{I}_0$

Answer

$\frac{\mathrm{I}_0}{3}$

Answer

$\frac{\mathrm{I}_0}{4}$

Answer

$\mathrm{\frac{I_0}{2}}$

Waves Optics

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