

InterviewSolution
Saved Bookmarks
1. |
A screen is placed `2m` away from the single narrow slit. Calculate the slit width if the first minimum lies `5 mm` on either side of the central maximum. Incident plane waves have a wavelength of `5000 Å`. |
Answer» Here, distance of the screen from the slit, `D = 2m, a = ?, x = 5mm = 5 xx 10^(-3)m`, `lambda = 5000 Å = 5000 xx 10^(-10)m` For the first secondary minima, ,brgt `sin theta = (lambda)/(a) = (x)/(D)` `:. a = (D lambda)/(x) = (2 xx 5000 xx 10^(-10))/(5 xx 10^(-3)) = 2 xx 10^(-4)m` |
|