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In Young's double slit experiment, the ratio of intensities at two points on a screen when waves from the two slits have a path difference of zero and $\frac{\lambda}{4}$ is
$\left[\cos 0^{\circ}=\sin 90^{\circ}=1, \sin 0^{\circ}=\cos 90^{\circ}=0\right]$
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$\left[\cos 0^{\circ}=\sin 90^{\circ}=1, \sin 0^{\circ}=\cos 90^{\circ}=0\right]$
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The correct answer is:
$2: 1$
$I=4 I_{0} \cos ^{2} \frac{\phi}{2}$
$\phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}=\frac{\pi}{2}$
$\frac{\phi}{2}=\frac{\pi}{4}$
$\frac{1}{I_{0}}=4 \cos ^{2} \frac{\pi}{4}=\frac{4}{2}=2: 1$
$\phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}=\frac{\pi}{2}$
$\frac{\phi}{2}=\frac{\pi}{4}$
$\frac{1}{I_{0}}=4 \cos ^{2} \frac{\pi}{4}=\frac{4}{2}=2: 1$
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