In Young's double slit experiment, the screen is moved 30 cm towards the slits. As a consequence, the fringe width of the pattern changes by 0.09 mm.…

$\displaystyle \beta_{1}=\frac{D \lambda}{d}$ $\displaystyle \beta_{2}=\frac{D_{2} \lambda}{d}$ $\displaystyle \beta_{1}-\beta_{2}=\frac{\lambda}{d}\left(D_{1}-D_{2}\right)$ $\displaystyle =\frac{\lambda}{d}\left[D-\left(D-30 \times 10^{-2}\right)\right]$ $\displaystyle \lambda=\frac{\left(\beta_{1}-\beta_{2}\right) d}{30 \times 10^{-2}}=\frac{9 \times 10^{-5} \times 2 \times 10^{-3}}{30 \times 10^{-2}}$ $\displaystyle \lambda=6 \times 10^{-7} \mathrm{~m}$