# in ydse two slits are 1mm apart and screen is placed 1m away from slits. calculate the fringe width when light of wave length 500 nm is used?ii) what should be the width of each slit in order to obtain 10maxima of the double slits pattern within central maximum of single slit pattern?

Arun
25758 Points
3 years ago

We need to obtain 10 maxima of double slit with in central maxima of single slit

For single slit, angular width of central maxima = $\frac{2\lambda }{b}$

or $\frac{y}{d}=\frac{2\lambda }{b}$or $y=\frac{2\lambda D}{b}$----------------------(1)

For interference pattern

Fringe width = $\frac{\lambda D}{d}$---------------------(2)

According to question

$\frac{2\lambda D}{b}=10.\frac{\lambda D}{d}$

$\therefore b=\frac{d}{5}=0.2mm$