In a Young's double slit experiment, the inte | vedclass

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(B) The intensity $I$ at any point in a Young's double slit experiment is given by $I = 4I_0 \cos^2(\frac{\phi}{2})$, where $\phi$ is the phase differ

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$3$

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

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