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In the Young's double slit experiment a monochromatic source of wavelength $\lambda$ is used. The intensity of light passing through each slit is $I_{0}$. The intensity of light reaching the screen $S_{C}$ at a point $P$, a distance $x$ from $O$ is given by (Take, $d< < D$ )
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The correct answer is:
$4 I_{0} \cos ^{2}\left(\frac{\pi d}{\lambda D} x\right)$
Path difference, $\Delta x=\frac{x d}{D}$
So, corresponding phase difference, $\phi=\frac{2 \pi}{\lambda}(\Delta x)$
$=\frac{2 \pi}{\lambda}\left(\frac{x d}{D}\right)$
The resultant intensity at $P$ is
$\begin{aligned}
I_{P} &=4 I_{0} \cos ^{2}\left(\frac{\phi}{2}\right) \\
&=4 I_{0} \cos ^{2}\left(\frac{2 \pi}{\lambda} \frac{x d}{D} \times \frac{1}{2}\right)=4 I_{0} \cos ^{2}\left(\frac{\pi d x}{\lambda D}\right)
\end{aligned}$
So, corresponding phase difference, $\phi=\frac{2 \pi}{\lambda}(\Delta x)$
$=\frac{2 \pi}{\lambda}\left(\frac{x d}{D}\right)$
The resultant intensity at $P$ is
$\begin{aligned}
I_{P} &=4 I_{0} \cos ^{2}\left(\frac{\phi}{2}\right) \\
&=4 I_{0} \cos ^{2}\left(\frac{2 \pi}{\lambda} \frac{x d}{D} \times \frac{1}{2}\right)=4 I_{0} \cos ^{2}\left(\frac{\pi d x}{\lambda D}\right)
\end{aligned}$
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