Light of wavelength 450nm is incident on a double slit of slit width 20um and sl

ID: 2110545 • Letter: L

Question

Light of wavelength 450nm is incident on a double slit of slit width 20um and slit separation of .2mm.

a) Find the maximum number of double-slit interference fringes one would expect to observe on a screen.

b) Find the maximum number of single-slit diffraction minima one would expect to observe on a screen.

c) Will any of the expected interference fringes be missing? If so, which is the first one expected to be missing?

d) Find the total number of bright interference fringes inside the central diffraction envelope.

Show work.

a) angle between consecutive maxima = wavelength/0.2mm = 0.0025 rad => no. of fringes = pi/0.0025 = 1396

b) angle between consecutive maxima = wavelength/20um = 0.0225 rad => no. of fringes = pi/0.0025 = 139

c) ratio of slit sepeartion/ slit width = 10... hence the tenth interefecnce maxima is missing

d) no of maxima inside envelope = 9 on upper side + central + 9 o lower side = 19

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