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(C) The resultant intensity in a Young's double slit experiment is given by $I = 4I_0 \cos^2\left(\frac{\Delta\phi}{2}\right)$,where $\Delta\phi$ is t

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$3$

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(C) The resultant intensity in a Young's double slit experiment is given by $I = 4I_0 \cos^2\left(\frac{\Delta\phi}{2}\right)$,where $\Delta\phi$ is the phase difference.Phase difference $\Delta\phi$ is related to path difference $\Delta x$ by $\Delta\phi = \frac{2\pi}{\lambda} \Delta x$.For path difference $\Delta x_1 = \frac{\lambda}{4}$,the phase difference is $\Delta\phi_1 = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2}$.Thus,$I_1 = 4I_0 \cos^2\left(\frac{\pi/2}{2}\right) = 4I_0 \cos^2\left(\frac{\pi}{4}\right) = 4I_0 \cdot \left(\frac{1}{\sqrt{2}}\right)^2 = 2I_0$.For path difference $\Delta x_2 = \frac{\lambda}{3}$,the phase difference is $\Delta\phi_2 = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3}$.Thus,$I_2 = 4I_0 \cos^2\left(\frac{2\pi/3}{2}\right) = 4I_0 \cos^2\left(\frac{\pi}{3}\right) = 4I_0 \cdot \left(\frac{1}{2}\right)^2 = I_0$.Therefore,$\frac{I_1 + I_2}{I_0} = \frac{2I_0 + I_0}{I_0} = 3$.

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