Paragraph I: In a Young's double slit experiment, each of the two slits A and B, as shown in the figure, are…

Paragraph I: In a Young's double slit experiment, each of the two slits A and B, as shown in the figure, are oscillating about their fixed center and with a mean separation of $0.8 \mathrm{~mm}$. The distance between the slits at time $t$ is given by $d=(0.8+0.04 \sin \omega t) \mathrm{mm}$, where $\omega=0.08 \mathrm{rad} \mathrm{s}^{-1}$. The distance of the screen from the slits is $1 \mathrm{~m}$ and the wavelength of the light used to illuminate the slits is $6000 Å$. The interference pattern on the screen changes with time, while the central bright fringe (zeroth fringe) remains fixed at point $O$. The $8^{\text {th }}$ bright fringe above the point $\mathrm{O}$ oscillates with time between two extreme positions. The separation between these two extreme positions, in micrometer $(\mu \mathrm{m})$, is ___________ .

Solution

$\begin{aligned} & y=n \cdot\left(\frac{\lambda D}{d}\right) \\ & \text { for } 8^{\text {th }} \text { fringe } \\ & y=8 \frac{\lambda D}{d} \\ & y_{\max }=8 \frac{\lambda D}{d_{\min }} \\ & y_{\min }=8 \frac{\lambda D}{d_{\max }} \\ & y_{\max }-y_{\min }=8 \lambda D\left[\frac{1}{d_{\min }}-\frac{1}{d_{\max }}\right] \\ & \lambda=6000 Å\end{aligned}$ $\begin{aligned} & \mathrm{D}=1 \mathrm{~m} \\ & \mathrm{~d}_{\max }=0.34 \mathrm{~mm} \\ & \mathrm{~d}_{\text {min }}=0.76 \mathrm{~mm} \\ & y_{\max }-y_{\min }=8 \times 6000 \times 10^{-10} \times 1\left[\frac{1}{0.76 \times 10^{-3}}-\frac{1}{0.84 \times 10^{-3}}\right] \\ & =8 \times 6 \times 10^{-4} \times\left[\frac{0.08}{0.76 \times 0.84}\right]=601.5 \mu \mathrm{m} \\ & \end{aligned}$ (answer range is from 598 to 602 ) `

Asked in: JEE Advanced 2024 (Paper 2)

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