Question

Light of wave length $6000A^{\circ }$ is used to obtain interference fringes of width $6mm$ in a Young's double slit experiment. Calculate the wave length of light required to obtain fringe of width $4mm$ when the distance between the screen and slits is reduced to half of its initial value.

Answer 1

Fringe width is given by $\beta =\frac{\lambda D}{d}$
In the first case, ${\lambda }_1=6000A^{\circ }=6000\times {10}^{-10}m$
${\beta }_1=6mm=0.006m$
$\therefore 0.006=6000\times {10}^{-10}\frac{D}{d}....\left(1\right)$
In the second case, the distance between the slits and the screen is $\frac{D}{2},{\beta }_2=4mm=0.004m$
$\therefore 0.004={\lambda }_2\times \frac{\frac{D}{2}}{d}....\left(2\right)$
Directing equations $\left(2\right)$ by $\left(1\right)$ we get
$\frac{0.004}{0.006}=\frac{{\lambda }_2}{2\times 6000\times {10}^{-10}}$
${\lambda }_2=\frac{0.004\times 2\times 6000\times {10}^{-10}}{0.006}$
${\lambda }_2=\frac{48\times {10}^{-10}}{0.006}$
${\lambda }_2=8000\times {10}^{-10}$
${\lambda }_2=8000A^{\circ }$.