In Young's double slit experiment the intensities at two points, for the path difference \(\frac{\lambda}{4}\) and \(\frac{\lambda}{3}\) ( \(\lambda=\) 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 \(\frac{I_1+I_2}{I_0}\) is equal to \(\left(\cos 60^{\circ}=0.5, \cos 45^{\circ}=\frac{1}{\sqrt{2}}\right)\)

Phase difference, \(\phi=\frac{2 \pi}{\lambda} \Delta l\)
For first point, \(\phi_1=\frac{2 \pi}{\lambda}\left(\frac{\lambda}{4}\right)\)
\(
\begin{aligned}
\therefore \quad \phi_1 & =\frac{\pi^c}{2} \\
\mathrm{I}_1 & =2 \mathrm{I}_0\left(1+\cos \phi_1\right) \\
\therefore \quad \mathrm{I}_1 & =2 \mathrm{I}_0 \quad \ldots .\left[\because \cos \phi_1=\cos (\pi / 2)=0\right]
\end{aligned}
\)
Similarly, for second point, \(\phi_2=\frac{2 \pi^e}{3}\)
\(
\begin{array}{ll}
\therefore \quad \mathrm{I}_2 & =2 \mathrm{I}_0\left(1+\cos \phi_2\right) \\
\therefore \quad \mathrm{I}_2 & =2 \mathrm{I}_0\left(1-\frac{1}{2}\right) \quad \ldots . .\left[\because \cos \left(\frac{2 \pi}{3}\right)=\frac{-1}{2}\right] \\
\therefore \quad \mathrm{I}_2 & =\mathrm{I}_0
\end{array}
\)
Hence, \(\frac{I_1+I_2}{I_0}=\frac{2 I_0+I_0}{I_6}=3\)