In a Young’s double slit experiment, the intensities at two...

As we know, resultant intensity is given by,

$I = 4{I}_{0}\cos^{2}$\left(\dfrac{\Deltaϕ}{2}\right)$and phase difference can be written as$∆ ϕ = \dfrac{2\pi}{\lambda} ∆ x$.</p><p>For first case:</p><p>${I}_{1} = 4{I}_{0}\cos^{2}\left(\dfrac{2\pi}{\lambda} \times \dfrac{\lambda}{4}\right) = 4{I}_{0}\cos^{2}\left(\dfrac{\pi}{4}\right) = 2{I}_{0}$</p><p>For second case:</p><p>${I}_{2} = 4{I}_{0}\cos^{2}\left(\dfrac{2\pi}{\lambda} \times \dfrac{\lambda}{3}\right) = 4{I}_{0}\cos^{2}\left(\dfrac{2\pi}{3}\right) = {I}_{0}$</p><p>Therefore, required value of$\dfrac{{I}_{1} + {I}_{2}}{{I}_{0}} = 3.