Given here: λ1=560×10−9m, β1=7.2×10−3m and β2=8.1×10−3m.
Let d be the width and D be the distance between slit and screen then fringe width, β=dλD.
Thus, from above, β∝λ
So, the wavelength of second light is λ2=β1β2λ1=89λ1
∴λ2=89×560=630nm.