Question

3. In a two-slit interference-diffraction pattern, there are 2dla -1 interference peaks within the central diffraction maximum. How many interference peaks are there within other diffraction maxima?

Answer 1

In a two-slit interference-diffraction For minimum diffraction a sin theta = m lambda (1) For maximum diffraction (a + b) sin theta = n lambda d sin theta = n lambda (2) [Because d = a + b] For central maximum a sin theta = lambda maxima d sin theta = n lambda Therefore n = d/a Therefore Total number of interference mass central diffraction max i.e: N = 2 (n - 1) + 1 = 2n - 2 + 1 N = 2n - 1 N = 2 d/a - 1 For the diffraction maxima By diffraction maxima By Dividing (1) and (2) i < i d/a = n/m (or) n = dm/a \r\n The n^th interference maxima will be atleast for the diffraction maxima Therefore Total number of interference maxima N = 2 (n - 1) + 1 = 2n - 2 + 1 N = -2 n - 1 N = 3 dm/a - 1