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In a Young's double slit experiment, the intensities at two points, for the path difference $\frac{\lambda}{4}$ and $\frac{\lambda}{3}$ ( $\lambda$ being the wavelength of light used) are $I_1$ and $I_2$ respectively. If $I_0$ denotes the intensity produced by each one of the individual slits, then $\frac{\mathrm{I}_1+\mathrm{I}_2}{\mathrm{I}_0}=\ldots \ldots$
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3
Resultant intensity in Young's double slit experiment
$\mathrm{I}=4 \mathrm{I}_0 \cos ^2\left(\frac{\Delta \phi}{2}\right)$
For path difference $\frac{\lambda}{4}$ phase difference,
$\Delta \phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}=\frac{\pi}{4}$
$\therefore \mathrm{I}_1=4 \mathrm{I}_0 \cos ^2\left(\frac{\pi}{4}\right)=2 \mathrm{I}_0$
For path difference $\frac{\lambda}{3}$
$I_2=4 I_0 \cos ^2\left(\frac{2 \pi}{\lambda} \times \frac{\lambda}{3}\right)=I_0$
$\therefore \frac{\mathrm{I}_1+\mathrm{I}_2}{\mathrm{I}_0}=3$
$\mathrm{I}=4 \mathrm{I}_0 \cos ^2\left(\frac{\Delta \phi}{2}\right)$
For path difference $\frac{\lambda}{4}$ phase difference,
$\Delta \phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}=\frac{\pi}{4}$
$\therefore \mathrm{I}_1=4 \mathrm{I}_0 \cos ^2\left(\frac{\pi}{4}\right)=2 \mathrm{I}_0$
For path difference $\frac{\lambda}{3}$
$I_2=4 I_0 \cos ^2\left(\frac{2 \pi}{\lambda} \times \frac{\lambda}{3}\right)=I_0$
$\therefore \frac{\mathrm{I}_1+\mathrm{I}_2}{\mathrm{I}_0}=3$
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