1. Condition for disappearance of fringes to the eye

   In Young’s double slit experiment, fringe width is $\beta = \frac{\lambda D}{d}$ where:

   • $\lambda = 600\text{ nm} = 600\times 10^{-9}\text{ m}$
   • $D = 1.0\text{ m}$
   • $d =$ slit separation

   The person is standing near the slits and observing the fringes on the screen. The fringes will not be resolvable by the eye when the angular separation of adjacent fringes becomes equal to the angular resolution of the eye.

2. Angular separation of adjacent fringes

   For a person near the slits, the screen is at distance $D$, so angular separation is approximately $\theta \approx \frac{\beta}{D}$

   Using $\beta = \frac{\lambda D}{d}$, $\theta \approx \frac{\lambda D/d}{D} = \frac{\lambda}{d}$

3. Resolution limit of the eye

   Given angular resolution of eye: $\theta_{\min} = \left(\frac{1}{60}\right)^\circ$

   Convert into radians: $\theta_{\min} = \frac{1}{60}\cdot \frac{\pi}{180} = \frac{\pi}{10800} \approx 2.91\times 10^{-4}\text{ rad}$

4. Set limiting condition

   At the limiting slit separation $d_0$, $\frac{\lambda}{d_0} = \theta_{\min}$

   Hence, $d_0 = \frac{\lambda}{\theta_{\min}}$

   Substituting values: $d_0 = \frac{600\times 10^{-9}}{2.91\times 10^{-4}}$

   $d_0 \approx 2.06\times 10^{-3}\text{ m}$

   $d_0 \approx 2.06\text{ mm}$

5. Closest option

   $d_0 \approx 2\text{ mm}$

   So the correct option is B.

6. Option check

   • A: $1\text{ mm}$ → too small
   • B: $2\text{ mm}$ → correct
   • C: $4\text{ mm}$ → too large
   • D: $3\text{ mm}$ → not as close as B

Final Answer: B: $2\text{ mm}$