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  $$\frac{\lambda}{3}\text{ is:}$$

Intensity after passing orthogonal polarizers $$P_1$$ and $$P_2$$:

$$I_1 = I_2 = \frac{I_0}{2}$$

Beams from $$s_1$$ and $$s_2$$ passing through $$P_3$$ at $$45^\circ$$ (using Malus's Law):

$$I_{s1} = I_1 \cos^2(45^\circ) \implies I_{s1} = \frac{I_0}{2} \times \frac{1}{2} = \frac{I_0}{4}$$

$$I_{s2} = I_2 \cos^2(45^\circ) \implies I_{s2} = \frac{I_0}{2} \times \frac{1}{2} = \frac{I_0}{4}$$

Since both beams pass through $$P_3$$, they have parallel transmission axes and can interfere.

$$\Delta x = \frac{\lambda}{3} \implies \phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{3}$$

Resultant intensity:

$$I_{\text{net}} = I_{s1} + I_{s2} + 2\sqrt{I_{s1}I_{s2}}\cos\phi$$

$$I_{\text{net}} = \frac{I_0}{4} + \frac{I_0}{4} + 2\left(\frac{I_0}{4}\right)\cos\left(\frac{2\pi}{3}\right) \implies I_{\text{net}} = \frac{I_0}{2} + \frac{I_0}{2}\left(-\frac{1}{2}\right) = \frac{I_0}{4}$$