Question

Light of wavelength 610 nm falls on a slit that is 3.80×10?3 mm wide. How far the first bright diffraction fringe is from the strong central maximum if the screen is 12.5 m away.

Answer 1

Path difference in constructive interference is given by:

d*sin = (m + 1/2)*

where, m = 0, 1, 2, 3,......

for first bright fringe from central maximum, m = 1

= wavelength of light = 610 nm = 610*10^-9 m

d = width of slit = 3.80*10^-3 mm = 3.80*10^-6 m

So,

sin = (1 + 1/2)*610*10^-9/(3.80*10^-6) = 0.2407895

= arcsin 0.2407895 = 13.93 deg

So distance between first bright fringe and central maximum will be:

y = L*tan = 12.5*tan 13.93 deg

y = 3.10 m

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