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  • In Young's double-slit experiment, let be the fringe width, and let <img alt="\mathrm{I}_{0}" src="https://entr

In Young's double-slit experiment, let \beta be the fringe width, and let \mathrm{I}_{0} be the intensity at the central bright fringe. At a distance \mathrm{x} from the central bright fringe, the intensity will be

Option: 1

\mathrm{I}_{0} \cos (\mathrm{x} / \beta)


Option: 2

I_{0} \cos ^{2}(x / \beta)


Option: 3

\mathrm{I}_{0} \cos ^{2}(\pi \times / \beta)


Option: 4

\frac{I_{0}}{4} \cos ^{2}\left(\frac{\pi x}{\beta}\right)


Answers (1)

best_answer

\mathrm{\Delta=\times \frac{d}{D}}
\mathrm{\therefore}  Phase difference \mathrm{=\phi=\frac{2 \pi}{\lambda} \Delta}

Let a = amplitude at the screen due to each slit.

\mathrm{\therefore \quad \mathrm{I}_{0}=\mathrm{k}(2 \mathrm{a})^{2}=4 \mathrm{ka}^{2}, where \: \mathrm{k}} is a constant.

For phase difference \mathrm{\phi}, amplitude \mathrm{=\mathrm{A}=2 \mathrm{a} \cos (\phi / 2)}.

\mathrm{Intensity, \mathrm{I}=\mathrm{kA} \mathrm{A}^{2}=\mathrm{k}\left(4 \mathrm{a}^{2}\right) \cos ^{2}(\phi / 2)=\mathrm{I}_{0} \cos ^{2}(\pi \Delta / \lambda)}

\mathrm{=\mathrm{I}_{0} \cos ^{2}\left(\frac{\pi}{\lambda} \cdot \frac{\mathrm{xd}}{\mathrm{D}}\right)=\mathrm{I}_{0} \cos ^{2}\left(\frac{\pi \mathrm{x}}{\beta}\right)}.
 

Posted by

sudhir kumar

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