Calculate the angular width of `2^(nd)` secondary maximum and fourth dark fringe in the diffraction pattern obtained due to a single slit of width `10^(-5) m ` illuminated by a wavelength of light `541` nm.

Explanation

(i) For `CI,(2n+1)(lambda)/(2)=d sin theta`
`sin theta+((2n+1)lambda)/(2d)+((2xx2+1)541xx10^(-9))/(2x10^(-5))=(2705xx10^(-4))/(2)=1352.5xx10^(-4)`
`sin theta=0.1352,` for small angle, `theta=0.1352 "rad" (~~ 8^@)`
angular width `=2 theta=0.2704 "rad".`
(ii) For `DI, nlambda=d sin theta`
`sin theta=(2164xx10^(-9))/(10^(-5))`
`sin theta=2164xx10^(-4)=0.2164` or `theta~~12^@`
angular width `=0.4328 "rad".`