Calculate the wavelength of light used in an interference...

Given: fringe with $\beta=0.03 \mathrm{ cm}, \mathrm{D}=1 \mathrm{ m}=100\mathrm{cm}$ Distance between images of the source $=0.8 \mathrm{ cm} .$ Image distance $v=80 \mathrm{ cm}$ Object distance $=u$ Using mirror formula, \begin{array}{l} \qquad \begin{array}{l}\frac{1}{v}+\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{60}+\frac{1}{u}=\frac{1}{16}\\ \Rightarrow \mathrm{u}=20 \mathrm{ cm} \end{array}$\\$\text{ Magnification, } \mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}=\frac{80}{20}=4 \\ \text{ Magnification } $$\begin{aligned} {\text{ distances between images of slits }}{\text{ distance between slits }}\\ =\frac{0.8}{\mathrm{ d}}=\frac{0.8}{\mathrm{ d}}=4 \Rightarrow \mathrm{d}=0.2 \mathrm{ cm} \end{aligned}$$$\\$\text{ Fringe width } \beta=\frac{\mathrm{D} \lambda}{\mathrm{d}} \end{array}$or,$\beta=\frac{100 \lambda}{0.2}=0.03 \times 10^{-2}$Therefore, wavelength of light used$\lambda=6000Ã…$