In YDSE, slits are `0.2mm` apart. The interference fringes for light of wavelength `6000 Å` are formed on a screen distant `1.5 m` from the slits. Calculate
(i) angular position of `3rd` maxima,
(ii) angular position of `5th` minima,
(iii) fringe width

Answer

(i) `0.009 rad`,
(ii) `0.0135 rad`
(iii) `4.5 mm`

Explanation

Here, `d = 0.2 mm = 2 xx 10^(-4)m`
`lambda = 6000 Å = 6 xx 10^(-7)m, D = 1.5 m`
(i) For angular position of 3rd maxima
`theta = (x_(3))/(D) = (n lambda)/(d) = (3 xx 6 xx 10^(-7))/(2 xx 10^(-4)) = 9 xx 10^(-3) rad`
(ii) For angular position of 5th minimum
`theta' = (x_(5))/(D) = (2n - 1)(lambda)/(d) = (2 xx 5 - 1) xx (6 xx 10^(-7))/(2 xx 2 xx 10^(-4))`
`= 13.5 xx 10^(-3) = 0.0135 rad`
(iii) Fringe width, `beta = (lambda D)/(d)`
`= (6 xx 10^(-7) xx 1.5)/(2 xx 10^(-4)) = 4.5 xx 10^(-3)m = 4.5 mm`