Question

In a double slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the plane of slits. If the screen is moved by $5 \times 10^{-2} m$ towards the slits, the change in fringe width is $3 \times 10^{-3} cm$. If the distance between the slits is $1 mm$, then the wavelength of the light will be ________$nm .$

Answer 1

$\beta=\frac{\lambda D }{ d }$
$\Delta \beta=\frac{\lambda}{ d } \Delta D$
$\lambda=\frac{\Delta \beta . d }{\Delta D }$
$=\frac{3 \times 10^{-5} \times 1 \times 10^{-3}}{5 \times 10^{-2}}$
$=60 \times 10^{-8}=600 \times 10^{-9} m$
$=600\, nm$