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

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

where:

• $\beta$ = separation between consecutive bright fringes,
• $\lambda$ = wavelength,
• $D$ = distance of screen from slits,
• $d$ = slit separation.

2. For the same experimental setup, $D$ and $d$ remain constant, so:

$\beta \propto \lambda$

Hence,

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

3. Given:

$\lambda_1 = 560\,\text{nm}$ $\beta_1 = 7.2\,\text{mm}$ $\beta_2 = 8.1\,\text{mm}$

So,

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

$\lambda_2 = 560 \cdot \frac{8.1}{7.2}$

4. Simplify the ratio:

$\frac{8.1}{7.2} = \frac{81}{72} = \frac{9}{8}$

Therefore,

$\lambda_2 = 560 \cdot \frac{9}{8}$

$\lambda_2 = 70 \cdot 9 = 630\,\text{nm}$

5. Final answer:

$\boxed{630}$