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JEE Advance - Physics (2024 - Paper 2 Online - No. 14)

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 $O$.

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 $O$.

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 _________ .
Answer
601.50

Explanation

$$\begin{aligned} & \mathrm{y}=\mathrm{n} .\left(\frac{\lambda \mathrm{D}}{\mathrm{d}}\right) \\ & \text { for } 8^{\text {th }} \text { fringe } \\ & \mathrm{y}=8 \frac{\lambda \mathrm{D}}{\mathrm{d}} \\ & \mathrm{y}_{\max }=8 \frac{\lambda \mathrm{D}}{\mathrm{d}_{\text {min }}} \\ & \mathrm{y}_{\text {min }}=8 \frac{\lambda \mathrm{D}}{\mathrm{d}_{\text {max }}} \\ & \mathrm{y}_{\text {max }}-\mathrm{y}_{\text {min }}=8 \lambda \mathrm{D}\left[\frac{1}{\mathrm{~d}_{\text {min }}}-\frac{1}{\mathrm{~d}_{\text {max }}}\right] \\ & \lambda=6000 \mathop A\limits^o\\ & \begin{aligned} & \mathrm{D}=1 \mathrm{~m} \\ & \mathrm{~d}_{\max }=0.34 \mathrm{~mm} \\ & \mathrm{~d}_{\min }=0.76 \mathrm{~mm} \\ & \mathrm{y}_{\text {max }}-\mathrm{y}_{\min }=8 \times 6000 \times 10^{-10} \times 1\left[\frac{1}{0.76 \times 10^{-3}}-\frac{1}{0.84 \times 10^{-3}}\right] \\ & \quad=8 \times 6 \times 10^{-4} \times\left[\frac{0.08}{0.76 \times 0.84}\right]=601.5 \mu \mathrm{m} \end{aligned} \end{aligned}$$

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