Question

Light of wavelength 610 nm falls on a slit 0.0572 mm wide.

(a) On a very large distant screen, how many totally dark fringes (indicating complete cancellation) will there be, including both sides of the central bright spot? Solve this problem without calculating all the angles! (Hint: What is the largest that sin?can be? What does this tell you is the largest that m can be?)
dark fringes

(b) At what angle will the dark fringe that is most distant from the central bright fringe occur?
±  °

Answer 1

(a) Let ' ' be the angular width of a fringe, 'd' be the distance between the two slits and be the wavelength of the light.
= / d = 610 * 10 ^-9 / 5.72 * 10 ^-5  = 0.010664335
Hence no. of fringes (n) = sin 90/sin
As is very small, sin ? ,
Therefore, n = 1/
n = d/ = 1 / = 1 / 0.010664335 = 93
Hence, the total number of dark fringes= 2n = 186

(b) farthest dark firnge at n = 93

d sin = n

sin = 93 * 610 * 10 ^-9 / 5.72 * 10 ^-5 = 0.9917

= 82.61283973