In a Young's double slit experiment, a laser light of 560nm...

Given here: ${\lambda}_{1} = 560 \times {10}^{- 9}m$, ${\beta}_{1} = 7 . 2 \times {10}^{- 3}m$ and ${\beta}_{2} = 8 . 1 \times {10}^{- 3}m$.

Let $d$ be the width and $D$ be the distance between slit and screen then fringe width, $\beta = \dfrac{\lambdaD}{d}$.

Thus, from above, $\beta ∝ \lambda$

So, the wavelength of second light is ${\lambda}_{2} = \dfrac{{\beta}_{2}}{{\beta}_{1}}{\lambda}_{1} = \dfrac{9}{8}{\lambda}_{1}$

$∴ {\lambda}_{2} = \dfrac{9}{8} \times 560 = 630nm .$