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  •  In YDSE, is the fringe width and <img alt="\mathrm{I_{0} \cos ^{2}(\pi y / \beta)}" src="https://entrancecorn

 In YDSE, \beta is the fringe width and \mathrm{I_{0} \cos ^{2}(\pi y / \beta)} is the intensity at a distance y from central bright fringe. In this case \mathrm{I}_{0} is

Option: 1

intensity of each wave


Option: 2

combined intensity of the two waves


Option: 3

 intensity at the central bright fringe


Option: 4

none of these.


Answers (1)

best_answer

\text{Phase difference }=\phi=\frac{2 \pi}{\lambda} \times \text{path difference }
\mathrm{=\frac{2 \pi}{\lambda}(\Delta x)}

Let \mathrm{a}= amplitude at screen due to each slit
\Rightarrow \quad \mathrm{I}_{0}= Intensity at central bright fringe
=\mathrm{k}(2 \mathrm{a})^{2}=4 \mathrm{ka}^{2}, \quad \mathrm{k}=\text { constant }

For phase difference \mathrm{\phi, \quad A=amplitude =2 a \cos (\phi / 2)}
\therefore \quad \mathrm{I}=\mathrm{kA}^{2}=\mathrm{k}(2 \mathrm{a})^{2} \cos ^{2}(\phi / 2)=4 \mathrm{ka}^{2} \cos ^{2}(\phi / 2)

=\mathrm{I}_{0} \cos ^{2}\left(\frac{\pi}{\lambda} \Delta \mathrm{x}\right)
=\mathrm{I}_{0} \cos ^{2}\left(\frac{\pi}{\lambda} \cdot \frac{\mathrm{yd}}{\mathrm{D}}\right)
\mathrm{=I_{0} \cos ^{2}\left(\frac{\pi y}{\beta}\right)}

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Ritika Harsh

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