JEE MAIN - Physics (2025 - 24th January Evening Shift - No. 2)
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
_24th_January_Evening_Shift_en_2_1.png)
Explanation
_24th_January_Evening_Shift_en_2_2.png)
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
$$\begin{aligned} & \Delta \phi=2 \mathrm{~K}\left(\frac{\Delta \mathrm{x}}{\lambda}\right)=\frac{2 \pi}{3} \\ & \therefore \mathrm{I}_{\text {res }}=\sqrt{\left(\frac{\mathrm{I}_0}{4}\right)^2+\left(\frac{\mathrm{I}_0}{4}\right)^2+2\left(\frac{\mathrm{I}_0}{4}\right)^2} \cos \frac{2 \pi}{3} \\ & =\frac{\mathrm{I}_0}{4} \end{aligned}$$
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