1. Given data

   • Slit separation: $d = 1\,\text{mm} = 10^{-3}\,\text{m}$
   • Wavelength: $\lambda = 632.8\,\text{nm}$
   • Screen distance: $D = 100\,\text{cm} = 1\,\text{m}$
   • दूरी of the bright fringe from central bright fringe: $y = 1.27\,\text{mm} = 1.27 \times 10^{-3}\,\text{m}$

2. Path difference at a point on screen

   In Young’s double slit experiment, the path difference at a point at transverse distance $y$ is

   $$\Delta = \frac{dy}{D}$$

3. Substitute the values

   $$\Delta = \frac{(10^{-3})(1.27 \times 10^{-3})}{1} = 1.27 \times 10^{-6}\,\text{m}$$

4. Convert into micrometre

   $$1.27 \times 10^{-6}\,\text{m} = 1.27\,\mu\text{m}$$

5. Check bright fringe condition

   For a bright fringe,

   $$\Delta = n\lambda$$

   Now,

   $$\frac{1.27\,\mu\text{m}}{632.8\,\text{nm}} = \frac{1270\,\text{nm}}{632.8\,\text{nm}} \approx 2$$

   so this corresponds to the second bright fringe, which is consistent.

6. Evaluate options

   • A: $1.27\,\mu\text{m}$ ✔️
   • B: $2.05\,\mu\text{m}$ ✖️
   • C: $2.87\,\text{nm}$ ✖️
   • D: $2\,\text{nm}$ ✖️

Therefore, the correct option is A.