In a two-slit setup, each slit is 0.020 mm wide. these apertures are illuminated by plane waves of yellow sodium light (? =589.6 nm). the resulting fraunhofer fringe pattern consists of 11 narrow bright fringes that gradually decrease in irradiance with distance from the central maximum. determine the separation between the slits.

so separation between slit is 0.12 mm

Explanation:

given data

slit wide = 0.020 mm

wavelength λ = 589.6 nm

fringe pattern n = 11

to find out

separation between the slits

solution

we know that for maxima we can say

d sinθ = m × λ   .........1

and m is mth order maxima

and for central maxima

number of bright fringe is here

N = 2m -1

11 = 2m - 1

m = 6

so here number of maxima on either side of central maxima will be m-1

that is = 5

because we know 6th maxima will be minimized by minima

so

condition for minima is

a sin(θ) = n × λ

now for 1st minima

a sin(θ)  =  λ

and

= m

d = ma

d = 6 × 0.02

d = 0.12 mm

so separation between slit is 0.12 mm

e/c, where e is its total energy