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In a Young's double slit experiment, the intensity of light at a point on the screen where the path difference between the interfering waves is $\lambda$, ( $\lambda$ being the wavelength of light used) is $\mathrm{I}$. The intensity at a point where the path difference is $\frac{\lambda}{4}$ will be (assume two waves have same amplitude)
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0.5I
$\mathrm{I}^{\prime}=\mathrm{I} \cos ^2 \phi / 2$
$\phi=\frac{2 \pi}{\lambda} \Delta \mathrm{x}, \quad \phi=\frac{2 \pi}{\lambda} \cdot \frac{\lambda}{4}=\frac{\pi}{2}$
$\frac{\phi}{2}=\frac{\lambda}{4}, \quad \mathrm{I}^{\prime}=\mathrm{I}^2 \cos ^2 \frac{\pi}{4}=\frac{\mathrm{I}}{2}$
$\phi=\frac{2 \pi}{\lambda} \Delta \mathrm{x}, \quad \phi=\frac{2 \pi}{\lambda} \cdot \frac{\lambda}{4}=\frac{\pi}{2}$
$\frac{\phi}{2}=\frac{\lambda}{4}, \quad \mathrm{I}^{\prime}=\mathrm{I}^2 \cos ^2 \frac{\pi}{4}=\frac{\mathrm{I}}{2}$
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