In a Young's Double slit experiment,the first | vedclass

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(C) The intensity at any point $P$ on the screen is given by $I_P = I_{\max} \cos^2(\phi/2)$,where $\phi$ is the phase difference.The phase difference

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first decreases and then increases

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(C) The intensity at any point $P$ on the screen is given by $I_P = I_{\max} \cos^2(\phi/2)$,where $\phi$ is the phase difference.The phase difference $\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \frac{yd}{D}$,where $y$ is the distance of point $P$ from the centre $O$,$d$ is the slit separation,and $D$ is the distance of the screen from the slits.At the centre $O$,$y = 0$,so $\phi = 0$ and $I_O = I_{\max}$.At point $P$,the first maxima is initially observed,so the path difference $\Delta x = \lambda$,which means $\phi = 2\pi$.As the screen is moved away,$D$ increases. Since $\phi = \frac{2\pi yd}{\lambda D}$,as $D$ increases,$\phi$ decreases from $2\pi$ towards $0$.As $\phi$ decreases from $2\pi$ to $0$,the value of $\cos^2(\phi/2)$ increases from $\cos^2(\pi) = 1$ to $\cos^2(0) = 1$,passing through $0$ at $\phi = \pi$ (the first minima).Therefore,the intensity $I_P$ first decreases (as it moves from maxima to minima) and then increases (as it moves from minima towards the central maxima). Thus,the ratio $I_P/I_O$ first decreases and then increases.

In a Young's Double slit experiment,the first maxima is observed at a fixed point $P$ on the screen. Now,the screen is continuously moved away from the plane of the slits. The ratio of the intensity at point $P$ to the intensity at point $O$ (centre of the screen) is:

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