Question 437: Wave Optics

Question

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 $O$.

Accepted Answer

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

Wave Optics

Integer / Numeric Answer Type