The width of fringe is 2 mm on the screen in a double slit experiment for the light of wavelength of 400 nm. The width of the fringe for the light of wavelength 600 nm will be:

In a double slit experiment, the fringe width is given by:

$$\beta = \frac{\lambda D}{d}$$

where $$\lambda$$ is the wavelength, $$D$$ is the distance to the screen, and $$d$$ is the slit separation. Since $$D$$ and $$d$$ remain constant, we have $$\beta \propto \lambda$$.

Now:

$$\frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1}$$

$$\beta_2 = \beta_1 \times \frac{\lambda_2}{\lambda_1} = 2 \times \frac{600}{400} = 2 \times 1.5 = 3 \text{ mm}$$

Hence, the correct answer is Option 4.