A screen is placed 50 cm from a single slit which is illuminated with light of wavelength 6000 Ã…. If the distance between the first and third minima in the diffraction pattern is 3.0 mm. The width of the slit is

Explanation

In case of diffraction at single slit, the position of minima is given by `d sin theta= n lamda` . where d is the aperture size and for small `theta : sin theta = theta =(y//D)`
`therefore d(y/D) = n lamda, i.e., y = D/d (n lamda)`
so that `y_3 - y_1 = D/d (3lamda -lamda) = D/d (2 lamda)` and hence
` d = (0.50xx(2 xx 6 xx 10^(-7)) )/(3 xx 10^(-3)) = 2 xx 10^(-4) m = 0.2 mm`