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In a Young double slit experiment, the wavelength of incident light is 6000 Å. The separation between slits $$S_1$$ and $$S_2$$ is 5 cm and the distance between slits plane and screen is 50 cm, as shown in the figure below. If the resultant intensity at P is equal to the intensity due to individual slits, the path difference between interfering waves is __________ Å.
$$I_0 = 4I_0 \cos^2\left(\frac{\phi}{2}\right)$$
$$\cos^2\left(\frac{\phi}{2}\right) = \frac{1}{4}$$
$$\frac{\phi}{2} = \frac{\pi}{3} \implies \phi = \frac{2\pi}{3}$$
$$\phi = \frac{2\pi}{\lambda} \cdot \Delta x$$
$$\frac{2\pi}{3} = \frac{2\pi}{\lambda} \cdot \Delta x$$
$$\Delta x = \frac{6000 \text{ \AA}}{3} = 2000 \text{ \AA}$$
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