A pair of thin horizontal slits in a vertical opaque screen is illuminated from

Question

A pair of thin horizontal slits in a vertical opaque screen is illuminated from the left with plane monochromatic light of wavelength 600 nm. The distance between the slits is 10 micrometers. To the right of the slits at a distance of 3 meters there is a vertical viewing screen.

a) How far above the center of the screen will the first dark fringe appear?
b) What is the smallest distance between a dark and a bright fringe?
c) If the material between the two slits were totally removed, what wil be the distance between the two dark fringes around the center of the resulting diffraction pattern?

Explanation / Answer

= 600 nm = 6.10-7 m

d = 10 m = 1.10-5 m

D = 3 m

A) y = D.(2n-1)/2d for first dark fringe subsitute n = 1

   y1d = D/2d = 0.09 m = 9 cm

B) for first bright fringe :

    y1b = D/d

so smallest distance y = y1b-y1d = D/2d = 0.09 m =9 cm

C) if the material removed this become a single slit :

    y = D/d = 0.18 m = 18 cm

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