A screen is placed `50 cm` from a single slit, which is illuminated with `6000 Å` light. If the distance between the first and third minima in the diffraction pattern is `3.00 mm`, what is the width of the slit ?

Explanation

In case of diffraction at single slit, the position of minima is given by ` a sin theta = n lambda ` . Where d is aperture size and for small `theta` :
` :. a ((y)/(D)) = n lambda , i.e., y = (D)/(a) (nlambda)`
So that , `y_(3) - y_(1) = (D)/(a) (3 lambda - lambda) = (D)/(a) (2 lambda) ` and hence,
`a = (050 xx (2 xx 6 xx 10^(-7)))/(3 xx 10^(-3)) = 2 xx 10^(-4)` m
`= 0 . 2 ` m m