Question

A double slit of separation 0.5 mm is illuminated by a parallel beam from a helium-neon laser that emits monochromatic light of wavelength 632.8 nm. Five meters beyond the slits is a screen. What is the separation of the interference fringes on the screen?

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Answer 1

Therefore, seperation of the interference fringes = $\frac{5 * 632.8 * 10^{-9}}{0.5 * 10^{-3}}$ = 6.328 mm

In Young's double slit experiment, (i) Fringe width of bright and dark fringe beta = D lambda/d where lambda = wavelength of wave D D = distance between slit and screen and d = distance between two slits Angular fringe width, theta = beta/D = lambda/d Separation of nth order bright fringe from central fringe y_n = Dn lambda/d n = 1, 2, 3 Therefore, separation of the interference fringes = 5 times 632.8 times 10^- 9/0.5 times 10^- 3 = 6.328 mm