Question

Light of wavelength 532 nm illuminates a pair of slits separated by a distance of d= 0.42 mm. An interference pattern is observed on a screen placed a distance L away (L>>d). You may use the small angle approximation for this problem. What is the distance L if the width of the central bright spot of the interference pattern is delta(y) = 1.9 cm? The answer is supposed to be 15 m but i cant figure out how to get that answer. Thanks!

Answer 1

Given:

Wavelength of light, λ = 532 nm = 532 x 10-9 m

Distance between the two slits, d = 0.42 mm = 0.42 x 10-3 m

The width of the central bright spot is nothing but the fringe width of the interference pattern which is given as ∆y = 1.9 cm = 0.019 m

The fringe width of the interference pattern, ∆y = λL/d where, L is the distance of the screen from the slits

L = ∆y d /λ = 0.019 x 0.42 x 10-3 /532 x 10-9 = 15 m

the distance of the screen from the slits, L = 15 m