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(A) The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ by the formula $\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$.Give

Vedclass · Accepted Answer

$3/4$

Vedclass · Answer

(A) The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ by the formula $\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$.Given $\Delta x = \lambda / 6$,we have $\Delta \phi = \frac{2 \pi}{\lambda} 	imes \frac{\lambda}{6} = \frac{\pi}{3}$.The intensity $I'$ at any point is given by $I' = I_0 \cos^2(\frac{\Delta \phi}{2})$,where $I_0$ is the maximum intensity.Substituting $\Delta \phi = \pi / 3$,we get $I' = I_0 \cos^2(\frac{\pi / 3}{2}) = I_0 \cos^2(\frac{\pi}{6})$.Since $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$,we have $I' = I_0 (\frac{\sqrt{3}}{2})^2 = I_0 	imes \frac{3}{4}$.Therefore,the ratio $I'/I_0 = 3/4$.

In Young's double-slit experiment,the intensity at a point where the path difference is $\lambda / 6$ is $I'$. If $I_0$ denotes the maximum intensity,then $I'/I_0$ is equal to

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