At two points P and Q on the screen in Young' | vedclass

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(B) Let $I_0$ be the intensity of each individual wave. The resultant intensity is given by $I = 2I_0 + 2I_0 \cos(\Delta \phi) = 4I_0 \cos^2(\frac{\De

Vedclass · Accepted Answer

$2:1$

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(B) Let $I_0$ be the intensity of each individual wave. The resultant intensity is given by $I = 2I_0 + 2I_0 \cos(\Delta \phi) = 4I_0 \cos^2(\frac{\Delta \phi}{2})$.At point $P$,the path difference $\Delta x = 0$,so the phase difference $\Delta \phi = 0$. The intensity $I_P = 4I_0 \cos^2(0) = 4I_0$.At point $Q$,the path difference $\Delta x = \frac{\lambda}{4}$,so the phase difference $\Delta \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2}$. The intensity $I_Q = 4I_0 \cos^2(\frac{\pi}{4}) = 4I_0 \cdot (\frac{1}{\sqrt{2}})^2 = 2I_0$.The ratio of intensities is $\frac{I_P}{I_Q} = \frac{4I_0}{2I_0} = 2:1$.

At two points $P$ and $Q$ on the screen in Young's double-slit experiment,waves from slits $S_1$ and $S_2$ have a path difference of $0$ and $\frac{\lambda}{4}$ respectively. The ratio of intensities at $P$ and $Q$ will be

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