Question

A single slit diffraction experiment performed with an argon laser of wavelength 454.6 nm produces a pattern on a screen with dark fringes (minima) separated by 10 mm. If we switch to a helium laser of wavelength 632.8 nm, without changing anything else in the setup, what will be the new separation between dark fringes? Select one: a. 13.9 mm b. 7.2 mm c. 15.2 mm d. 17.3 mm e. 6.1 mm

Answer 1

Warelength of argen laser =+ = 454.6 nm Warelength of helium laser = x₂ = 632.8mm fringe separation for t = 8, 10 mm fringe separation for ₂ = 32 fringe separation is given by ß= XD Where a C D d warelength of light separation between slit and screen slich wiath B, = X, D ... o R2 = 4.3D ---- @ For X, 8 for tot dividing @ uy , we get gate cinema >> Po ashtu si = = 632.8 mm (10mm) - mm) 632.8 nm 454. 6 nm B = 13.9 mm > ß2= 1. 39 x lomm = Answer (a)