In a biprism experiment, the slit separation is \(1 \mathrm{~mm}\). Using monochromatic light
of wavelength \(5000 Å\), an interference pattern is obtained on the screen. Where
should the screen be moved, so that the change in fringe width is \(12 \cdot 5 \times 10^{-5} \mathrm{~m}\) ?

\(\mathrm{d}=1 \times 10^{-3} \mathrm{~m}\)
\(\lambda=5000 Å=5 \times 10^{-7} \mathrm{~m}\)
\(\beta_{2}-\beta_{1}=12.5 \times 10^{-5} \mathrm{~m}\)
\(\mathrm{D}_{2}-\mathrm{D}_{1}=?\)
\(\beta_{1}=\frac{\lambda \mathrm{D}_{1}}{\mathrm{~d}} \quad \beta_{2}=\frac{\lambda \mathrm{D}_{2}}{\mathrm{~d}}\)
\(\beta_{2}-\beta_{1}=\frac{\lambda}{\mathrm{d}}\left(\mathrm{D}_{2}-\mathrm{D}_{1}\right)\)
\(\therefore \quad\left(\mathrm{D}_{2}-\mathrm{D}_{1}\right)=\frac{\mathrm{d}}{\lambda}\left(\beta_{2}-\beta_{1}\right)\)
\(\quad\left(\mathrm{D}_{2}-\mathrm{D}_{1}\right)=\frac{10^{-3}}{5 \times 10^{-7}} \times 12.5 \times 10^{-5}=\frac{12.5}{50}=25 \mathrm{~cm}\)