QJEEAdv24Ph7b0c: JEE Advanced 2024 Physics - Class 12 JEEAdv24Ph7b0c Solution: The bright fringe above the point oscillates with time between…

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