Question
In interference experiment the path difference between two interfering waves at a point A on the screen is $\lambda/3$, where $\lambda$ is the wavelength of these waves, and at another point B the path difference is $\lambda/6$. The ratio of intensities at points A and B is
$\_\_\_\_$。
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Solution
Sol. [Key Concept: Intensity $I = 4I_{0}\cos^{2}\frac{\Delta\varphi}{2}$ with $\Delta\varphi = \frac{2\pi}{\lambda} \times (path)$; take the ratio for paths $\lambda/3$ and $\lambda/6$.]
$${\ \Delta\phi_{A} = \frac{2\pi}{\lambda} \times \frac{\lambda}{3} = \frac{2\pi}{3} }{\therefore I_{A} = I + I + 2\sqrt{I} \cdot \sqrt{I} \cdot cos\left( \frac{2\pi}{3} \right) = I }{\Delta\phi_{B} = \frac{2\pi}{\lambda} \times \frac{\lambda}{6} = \frac{\pi}{3} }{\therefore I_{B} = I + I + 2\sqrt{I} \cdot \sqrt{I} \cdot cos\left( \frac{\pi}{3} \right) = 3I }{\frac{I_{A}}{I_{B}} = \frac{I}{3I} = \frac{1}{3} }$$
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