The maximum intensity in Young's double slit | vedclass

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Question

(A) Let $P$ be a point in front of one slit where the intensity is to be calculated. From the figure,it is clear that the distance of point $P$ from t

Vedclass · Accepted Answer

$\frac{I_0}{2}$

Vedclass · Answer

(A) Let $P$ be a point in front of one slit where the intensity is to be calculated. From the figure,it is clear that the distance of point $P$ from the central axis is $x = \frac{d}{2}$.The path difference between the waves reaching at point $P$ is given by:$\Delta = \frac{xd}{D} = \frac{(\frac{d}{2})d}{10d} = \frac{d}{20}$.Given $d = 5\lambda$,substituting this value:$\Delta = \frac{5\lambda}{20} = \frac{\lambda}{4}$.The corresponding phase difference $\phi$ is:$\phi = \frac{2\pi}{\lambda} 	imes \Delta = \frac{2\pi}{\lambda} 	imes \frac{\lambda}{4} = \frac{\pi}{2}$.The resultant intensity $I$ at point $P$ is given by:$I = I_0 \cos^2(\frac{\phi}{2}) = I_0 \cos^2(\frac{\pi}{4}) = I_0 	imes (\frac{1}{\sqrt{2}})^2 = \frac{I_0}{2}$.

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