InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Express angular width of central maximum in terms of wavelength. |
|
Answer» SOLUTION :For a diffraction minima `n lambda=d sin theta`, to find the WIDTH of the central MAXIMUM, put `n=1` then`sin theta=(lambda)/(d):theta=sin^(-1)((lambda)/(d))`. Here `'theta'` is semi angular width. For a TOTAL angular width `theta'=20=2 sin^(-1)((lambda)/(d))` |
|