1. In Young’s double slit experiment, the position of the $n$th bright fringe from the central maximum is $y_n=n\beta=n\frac{\lambda D}{d}$ where $\beta$ is the fringe width.

2. For the first source:

• Wavelength $\lambda_1=600\,\text{nm}$
• Position of 10th bright fringe: $y_{10}=10\,\text{mm}$

So, $y_{10}\propto \lambda$ for the same experimental setup ($D$ and $d$ unchanged).

3. When the wavelength is changed to $\lambda_2=660\,\text{nm}$ the new position of the same 10th bright fringe will scale in the same ratio: $\frac{y_2}{y_1}=\frac{\lambda_2}{\lambda_1}$

Thus, $y_2=10\times \frac{660}{600}$ $y_2=10\times 1.1=11\,\text{mm}$

4. Therefore, the distance of the 10th bright fringe from the central maximum is $\boxed{11\,\text{mm}}$