Wave Optics

In Young's double slit experiment using monochromatic light with a wavelength of $6000\ \text{\AA}$, the initial fringe width ( $\beta_1$ ) is 3 mm . We are tasked with determining the new fringe width when the distance between the screen and the slits is increased by $50\%$, and the distance between the slits is decreased by $10\%$.Given:Wavelength, $\lambda=6000\ \text{\AA}$Initial fringe width, $\beta_1=3\ \mathrm{mm}$The formula for fringe width $(\beta)$ in Young's experiment is:$\beta=\;\frac{D\lambda}{d}$where $D$ is the distance between the screen and the slits, and $d$ is the distance between the slits. From the given information:$3=\;\frac{D\times 6000}{d}$Now, recalculating for the changed conditions:The new distance between the screen and slits, $D'$, when increased by $50\%$ :$D'=D+\;\frac{D\times 50}{100}=1.5D$The new distance between the slits, $d'$, when decreased by $10\%$ :$d'=d-\;\frac{d\times 10}{100}=0.9d$Now, the new fringe width, $\beta'$, is given by:$\beta'=\;\frac{D'\lambda}{d'}=\;\frac{1.5\times D\times 6000}{0.9\times d}$Substituting Eq. (1) into the equation:$\beta'=\;\frac{1.5}{0.9}\times 3$ Calculating the above expression:$\beta'=\;\frac{45}{9}=5\ \mathrm{mm}$Thus, the new fringe width is 5 mm .