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  • Consider a usual set up of Young's double slit experiment with slits of equal intensity as shown in the figure. Take <img alt="\mathrm{'O'}" src="https://entrancecorner.oncodecogs.com/gif.latex

Consider a usual set up of Young's double slit experiment with slits of equal intensity as shown in the figure. Take \mathrm{'O'} as origin and the \mathrm{Y} axis as indicated. If average intensity between \mathrm{y}_{1}=$ $\frac{\lambda D}{4 d}$ and $y_{2}=+\frac{\lambda D}{4 d} equals \mathrm{'n'} times the intensity of maxima, then \mathrm{'n'} equals

Option: 1

\frac{1}{2}\left(1+\frac{2}{\pi}\right)


Option: 2

2\left(1+\frac{2}{\pi}\right)


Option: 3

\left(1+\frac{2}{\pi}\right)


Option: 4

\frac{1}{2}\left(1-\frac{2}{\pi}\right)


Answers (1)

best_answer

Phase difference corresponding to \mathrm{y}_{1}=-\pi / 2 and that for \mathrm{y}_{2}=+\pi / 2

\therefore  Average intensity between \mathrm{y}_{1}$ and $\mathrm{y}_{2}=\int_{-\pi / 2}^{\pi / 2} \frac{\mathrm{I}+\mathrm{I}+2 \sqrt{\| l} \cos \phi \mathrm{d} \phi}{\pi}

\mathrm{=\frac{21}{\pi} \int_{-\pi / 2}^{\pi / 2}(1+\cos \phi) \mathrm{d} \phi=\frac{2 \mid(\pi+2)}{\pi}}

\mathrm{\mathrm{I}_{\max }=4 \mathrm{I}}
\mathrm{\therefore} The required ratio \mathrm{=\frac{1}{2}(1+2 / \pi)}.

Posted by

seema garhwal

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