In Young's double slit experiment, the slits are separated by 0.6 mm and screen is placed at a distance of 1.2 m from slit. It is observed that the tenth bright fringe is at a distance of 8.85 mm from the third dark fringe on the same side. The wavelength of light used is

For \(\mathrm{n}^{\text {th }}\) bright fringe, \(\mathrm{y}_{\mathrm{n}}=\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}}\)
\(\therefore \quad \mathrm{y}_{10}=\frac{10 \times \lambda \times 1.2}{0.6 \times 10^{-3}}=\left(20 \times 10^3\right) \lambda...(i)\)
For \(\mathrm{n}^{\text {th }}\) dark fringe, \(y_n^{\prime}=\frac{(2 n-1) \lambda D}{2 d}\)
\(\therefore \quad y_3^{\prime}=\frac{5 \lambda \times 1.2}{2 \times\left(0.6 \times 10^{-3}\right)}=\left(5 \times 10^3\right) \lambda...(ii)\)
Given,
\(\begin{aligned}
& y_{10}-y_3^{\prime}=8.85 \times 10^{-3} \\
& \left(20 \times 10^3\right) \lambda-\left(5 \times 10^3\right) \lambda=8.85 \times 10^{-3}
\end{aligned}\)
...[From(i) and (ii)]
\(\begin{aligned}
& \lambda=\frac{8.85 \times 10^{-3}}{15 \times 10^3}=5.9 \times 10^{-7} \mathrm{~m} \\
& \lambda=5900 Å
\end{aligned}\)