1. Fringe width in YDSE

For Young's double slit experiment, eta = \frac{\lambda D}{d} where

• $\lambda = 5000\,\text{\AA} = 5000 \times 10^{-10}\,\text{m} = 5 \times 10^{-7}\,\text{m}$
• $d = 1.0\,\text{mm} = 10^{-3}\,\text{m}$
• $D = 1.0\,\text{m}$

So, $\beta = \frac{(5 \times 10^{-7})(1)}{10^{-3}} = 5 \times 10^{-4}\,\text{m}$

2. Condition for intensity to become half of maximum

In YDSE, intensity at a point is $I = I_{\max} \cos^2\left(\frac{\phi}{2}\right)$

For first time, $I = \frac{I_{\max}}{2}$ So, $\cos^2\left(\frac{\phi}{2}\right) = \frac{1}{2}$

Thus, $\frac{\phi}{2} = \frac{\pi}{4}$ for the first time. Hence, $\phi = \frac{\pi}{2}$

Now phase difference is related to path difference $\Delta$ by $\phi = \frac{2\pi \Delta}{\lambda}$ So, $\frac{2\pi \Delta}{\lambda} = \frac{\pi}{2}$ $\Delta = \frac{\lambda}{4}$

3. Convert path difference to position on screen

In YDSE, $\Delta = \frac{dy}{D}$ Thus, $\frac{dy}{D} = \frac{\lambda}{4}$ $y = \frac{\lambda D}{4d} = \frac{\beta}{4}$

So, $y = \frac{5 \times 10^{-4}}{4} = 1.25 \times 10^{-4}\,\text{m}$

4. Write in the required form

We need $y = \underline{\hspace{1cm}} \times 10^{-6}\,\text{m}$

Since $1.25 \times 10^{-4} = 125 \times 10^{-6}$

Therefore, the required integer is $\boxed{125}$