QJEEAdv24Phc9f1: JEE Advanced 2024 Physics - Class 12 JEEAdv24Phc9f1 Solution: The maximum speed in at which the bright fringe will move is…

$\begin{aligned}& \mathrm{y}=\mathrm{n} \cdot \frac{\lambda \mathrm{D}}{\mathrm{d}} \\& \mathrm{v}=\frac{\mathrm{dy}}{\mathrm{dt}}=-\mathrm{n} \cdot \frac{\lambda \cdot \mathrm{d}}{\mathrm{d}^2} \cdot \frac{\mathrm{d}(\mathrm{d})}{\mathrm{dt}} \\& \mathrm{d}=0.8+0.04 \sin \omega \mathrm{t} \\& \frac{\mathrm{d}(\mathrm{d})}{\mathrm{dt}}=0.04 \omega \cos \omega \mathrm{t} \\& \text { for } \mathrm{v} \rightarrow \max \Rightarrow \frac{\mathrm{d}(\mathrm{d})}{\mathrm{dt}} \rightarrow \max \\& \text { For } \frac{\mathrm{d}(\mathrm{d})}{\mathrm{dt}} \rightarrow \max \\& \cos \omega \mathrm{t}=1 \Rightarrow \sin \omega \mathrm{t}=0 \\& \Rightarrow\left(\frac{\mathrm{d}(\mathrm{d})}{\mathrm{dt}}\right)_{\max }=0.04 \\& \Rightarrow \mathrm{d}=0.8 \mathrm{~mm} \\& \mathrm{v}_{\max }=\frac{8 \times 6000 \times 10^{-10} \times 1 \times 0.04 \times 0.08}{0.8 \times 0.8 \times 10^{-6} \times 10^{-3}}=24 \mu \mathrm{m} / \mathrm{s}\end{aligned}$