Light of wavelength 441.1 nm falls on two slits spaced 0.29 mm apart. What is the required distance from the slits to the screen if the spacing between the first and second dark fringes is to be 4.0 mm?

R = 0.131 m

Explanation: The formulae for the distance between a fringe and the first (center) is given by

y = R×mλ/d

Where y = distance between first and nth fringe = 4mm = 4×10^-3m

λ = wavelength of light = 441.1nm = 441.1×10^-9m

R = distance between slits and screen =?

d = distance between slits = 0.29mm = 2.9×10^-4m

4×10^-3 = R ×2 ×441.4×10^-9/ 2.9×10^-5

Hence R = (4×10^-3) ×(2.9×10^-5)/2×441.4×10^-9

R = 1.16 × 10^-7/8.828×10^-7

R = 1.16/8.828

R = 0.131 m

area of focus