Question

5. Light of wavelength 632.8 nm shines upon a system of 5 narrow slits. Each slit is o.o04 mm wide and the slits are separated by o.12 mm a) If four of the slits are blocked, what is the angle from the centerline to the first point of zero intensity (dark spot)? b) With all five slits uncovered, find the angle (central maximum so) to the first point of zero intensity. Draw a phasor diagram at that point. c) Sketch the resulting intensity pattern when five slits are uncovered. Include labels where appropriate.

Answer 1

lambda = 632.8 nm b = 0.004 mm = 4 mu m c = 0.12 mn c = 120 mu m (a) b sin theta = n lambda For First minima n = 1 0.004 mm sin theta = 632.8 nm sin theta = 0.1582 theta = 9.1024 degree \r\n (b) when all 5 - slit open there is also interference patter n = 5 and First minimum For Interface 2c sin theta = lambda sin theta = 632.8 times 10^-9/2 times 120 times 10^-6 sin theta = 2.636 times 10^-3 theta = 0.1510 degree \r\n (c) N = 1 N = 2 N = 3 N = 4 N = 5