The 8^th bright fringe above the point O oscillates with time between two extreme positions. The separation between these two extreme positions, in micrometer (𝜇m), is ______.

The correct answer is 601.5.

Question Details and Parameters:
Based on the Young's double-slit experiment (YDSE) setup, the parameters are:
- The wavelength of the light: $\lambda =6000\text{Å}=6\times {10}^{-7}\text{m}$
- The distance between the slits and the screen: $D=1\text{m}$
- The separation between the slits varies with time as: $d\left(t\right)=(0.8+0.04sin\omega t)\text{mm}$

Step-by-Step Derivation:

1. Determine the limits of the slit separation:
The function $sin\omega t$ oscillates between $-1$ and $1$. Therefore, the slit separation $d\left(t\right)$ oscillates between a minimum value $d_{\text{min}}$ and a maximum value $d_{\text{max}}$:
$d_{\text{min}}=0.8-0.04=0.76\text{mm}=0.76\times {10}^{-3}\text{m}$

$d_{\text{max}}=0.8+0.04=0.84\text{mm}=0.84\times {10}^{-3}\text{m}$

2. Fringe Position Formula:
The position $y_n$ of the $n$-th bright fringe from the central maximum is given by:
$y_n=\frac{n\lambda D}{d}$

3. Extreme Positions of the 8th Bright Fringe (n = 8):
The maximum distance from the central maximum occurs when the slit separation is at its minimum:
$y_{\text{8,max}}=\frac{8\lambda D}{d_{\text{min}}}$

The minimum distance from the central maximum occurs when the slit separation is at its maximum:
$y_{\text{8,min}}=\frac{8\lambda D}{d_{\text{max}}}$

4. Calculate the Separation Between the Extreme Positions:
The separation $\Delta y$ between these two extreme positions is:
$\Delta y=y_{\text{8,max}}-y_{\text{8,min}}=8\lambda D(\frac{1}{d_{\text{min}}}-\frac{1}{d_{\text{max}}})$

Substitute the given values into the equation:
$\Delta y=8\times (6\times {10}^{-7}\text{m})\times 1\text{m}\times (\frac{1}{0.76\times {10}^{-3}\text{m}}-\frac{1}{0.84\times {10}^{-3}\text{m}})$

$\Delta y=\frac{48\times {10}^{-7}}{{10}^{-3}}(\frac{1}{0.76}-\frac{1}{0.84})\text{m}$

$\Delta y=48\times {10}^{-4}\left(\frac{0.84-0.76}{0.76\times 0.84}\right)\text{m}$

$\Delta y=48\times {10}^{-4}\left(\frac{0.08}{0.6384}\right)\text{m}$

$\Delta y=48\times {10}^{-4}\times 0.125313\text{m}$

$\Delta y\approx 6.015\times {10}^{-4}\text{m}$

5. Convert to micrometers (μm):
Since $1\text{m}={10}^6\text{μm}$:
$\Delta y\approx 6.015\times {10}^{-4}\times {10}^6\text{μm}=601.5\text{μm}$

Thus, the separation between the two extreme positions of the 8th bright fringe is 601.5 μm.