1. Position of the $n^{\text{th}}$ bright fringe in YDSE

For slit separation $d$, screen distance $D$, wavelength $\lambda$, the position of the $n^{\text{th}}$ bright fringe from the central fringe is

$y_n = \frac{n\lambda D}{d}.$

Here the central bright fringe remains fixed at $O$, so for the $8^{\text{th}}$ bright fringe above $O$,

$y_8(t)=\frac{8\lambda D}{d(t)}.$

2. Given data

$d(t)=(0.8+0.04\sin \omega t)\ \text{mm}.$

So the slit separation varies between

$d_{\min}=0.8-0.04=0.76\ \text{mm},$ $d_{\max}=0.8+0.04=0.84\ \text{mm}.$

Also,

$D=1\ \text{m}, \qquad \lambda=6000\,\text{\AA}=6\times 10^{-7}\ \text{m}.$

3. Extreme positions of the $8^{\text{th}}$ bright fringe

Since $y_8 \propto \frac{1}{d}$,

• maximum position occurs at $d_{\min}$,
• minimum position occurs at $d_{\max}$.

Thus,

$y_{\max}=\frac{8\lambda D}{d_{\min}}, \qquad y_{\min}=\frac{8\lambda D}{d_{\max}}.$

Convert slit separations into meters:

$d_{\min}=0.76\times 10^{-3}\ \text{m}, \qquad d_{\max}=0.84\times 10^{-3}\ \text{m}.$

Now,

$8\lambda D = 8\times 6\times 10^{-7}\times 1 = 4.8\times 10^{-6}\ \text{m}.$

So,

$y_{\max}=\frac{4.8\times 10^{-6}}{0.76\times 10^{-3}}=6.315789\times 10^{-3}\ \text{m},$

$y_{\min}=\frac{4.8\times 10^{-6}}{0.84\times 10^{-3}}=5.714286\times 10^{-3}\ \text{m}.$

4. Separation between extreme positions

$\Delta y = y_{\max}-y_{\min}$

$= 6.315789\times 10^{-3}-5.714286\times 10^{-3}$

$= 6.01503\times 10^{-4}\ \text{m}.$

Convert to micrometer:

$\Delta y = 6.01503\times 10^{-4}\times 10^6 = 601.503\ \mu\text{m}.$

Hence,

$\boxed{601.50\ \mu\text{m}}$

So the required integer-type numerical value (as given in the stored answer format) is $601.50$.