The maximum speed in 𝜇m/s at which the 8^th bright fringe will move is __________.

The correct answer is 24.

1. Identify the given parameters from the Young's Double Slit Experiment (YDSE):
- Mean slit separation: $d_0=0.8\text{mm}$
- Time-varying slit separation: $d\left(t\right)=(0.8+0.04\sin (\omega t\left)\right)\text{mm}=(0.8+0.04\sin (\omega t\left)\right)\times {10}^{-3}\text{m}$
- Angular frequency of oscillation: $\omega =0.08{\text{rad s}}^{-1}$
- Distance to the screen: $D=1\text{m}$
- Wavelength of the light source: $\lambda =6000\text{Å}=6��{10}^{-7}\text{m}$
- Fringe order of interest: $m=8$

2. Express the position of the 8^th bright fringe:
The position of the $m^{\text{th}}$ bright fringe on the screen relative to the central maximum is given by:
$y_m\left(t\right)=\frac{m\lambda D}{d\left(t\right)}$
For the 8^th bright fringe ($m=8$):
$y_8\left(t\right)=\frac{8\lambda D}{d\left(t\right)}$

3. Differentiate to find the velocity of the fringe:
Differentiating $y_8\left(t\right)$ with respect to time $t$ using the chain rule gives:
$v\left(t\right)=\frac{d{y}_8}{dt}=-\frac{8\lambda D}{{\left[d\right(t\left)\right]}^2}·\frac{d\left[d\right(t\left)\right]}{dt}$

Since $d\left(t\right)=(0.8+0.04\sin (\omega t\left)\right)\times {10}^{-3}\text{m}$, its derivative with respect to time is:
$\frac{d\left[d\right(t\left)\right]}{dt}=0.04\omega \cos \left(\omega t\right)\times {10}^{-3}\text{m/s}$

Substitute this back into the velocity equation:
$\left|v\right(t\left)\right|=\frac{8\lambda D}{{\left[d\right(t\left)\right]}^2}·0.04\omega |\cos (\omega t\left)\right|\times {10}^{-3}$

4. Find the maximum speed:
Because the amplitude of the separation variation ($0.04\text{mm}$) is extremely small compared to the average slit distance ($0.8\text{mm}$), the maximum value of velocity occurs very close to the instant when $\cos \left(\omega t\right)\approx 1$ and $\sin \left(\omega t\right)\approx 0$.
Substituting these values:
$v_{\text{max}}\approx \frac{8·(6\times {10}^{-7})·1}{{(0.8\times {10}^{-3})}^2}·(0.04\times {10}^{-3})·\left(0.08\right)$

Simplify the terms:
$v_{\text{max}}\approx \frac{4.8\times {10}^{-6}}{0.64\times {10}^{-6}}·3.2\times {10}^{-6}$
$v_{\text{max}}\approx 7.5\times 3.2\times {10}^{-6}\text{m/s}$
$v_{\text{max}}\approx 24\times {10}^{-6}\text{m/s}=24\mu \text{m/s}$