1. In Young’s double slit experiment, the fringe width is

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

where:

• $\lambda$ = wavelength,
• $D$ = distance of screen from slits,
• $d$ = slit separation.

2. Initially,

$\beta_1 = 0.5\ \text{mm}, \quad \lambda_1 = 5000\,\mathring{A}$

3. In the new setup:

• wavelength changes to $\lambda_2 = 6000\,\mathring{A}$
• slit separation is doubled: $d_2 = 2d_1$

4. Using proportionality from

$\beta \propto \frac{\lambda}{d}$

we get

$\frac{\beta_2}{\beta_1} = \frac{\lambda_2/d_2}{\lambda_1/d_1}$

Substitute $d_2 = 2d_1$:

$\frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1}\cdot\frac{d_1}{2d_1} = \frac{\lambda_2}{2\lambda_1}$

$\frac{\beta_2}{\beta_1} = \frac{6000}{2\times 5000} = \frac{6000}{10000} = 0.6$

5. Therefore,

$\beta_2 = 0.6\beta_1 = 0.6 \times 0.5\ \text{mm} = 0.3\ \text{mm}$

6. Hence, the new fringe width is

$\boxed{0.3\ \text{mm}}$

So the correct option is D.